剑指offer

2018-2-25

面试题7：重建二叉树

根据前序遍历和中序遍历重建二叉树。

思路：根据前序遍历第一个作为根节点，在中序遍历中划分为左子树和右子树。递归处理。

边界判断需要小心。

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
12         if (pre == null || in == null || pre.length != in.length) {
13             return null;
14         }
15         TreeNode res = reConstr(pre, 0, pre.length - 1, in, 0, in.length - 1);
16         return res;
17     }
18     private TreeNode reConstr(int[] pre,int startPre, int endPre, int[] in, int startIn,int endIn) {
19         if ((startPre > endPre) || (startIn > endIn)) {
20             return null;
21         }
22         int node = pre[startPre];
23         int pos = startIn;
24         for (int i = startIn; i <= endIn; i++) {
25             if (in[i] == node) {
26                 pos = i;
27                 break;
28             }
29         }
30         TreeNode res = new TreeNode(pre[startPre]);
31         int countLeft = pos - startIn;
32         res.left = reConstr(pre, startPre + 1, startPre + countLeft,in, startIn, pos - 1);
33         res.right = reConstr(pre,startPre + countLeft + 1, endPre,in,pos + 1, endIn);
34         return res;
35     }
36 }

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面试题8：二叉树的下一个节点

给定二叉树和其中一个节点，找出中序遍历的下一个节点。（有指向父节点的指针）

思路：思路要清晰

中序遍历：左-中-右

①判空

②如果该节点存在右子树，则返回右子树

③不存在右子树，则往上找，直到找到一个点，是其父节点的左节点，返回该父节点。

④找到根节点都不存在这样的父节点，则说明该点是最后一个点，返回空。

 1 /*
 2 public class TreeLinkNode {
 3     int val;
 4     TreeLinkNode left = null;
 5     TreeLinkNode right = null;
 6     TreeLinkNode next = null;
 7 
 8     TreeLinkNode(int val) {
 9         this.val = val;
10     }
11 }
12 */
13 public class Solution {
14     public TreeLinkNode GetNext(TreeLinkNode pNode)
15     {
16         if (pNode == null) {
17             return null;
18         }
19         if (pNode.right != null) {
20             TreeLinkNode temp = pNode.right;
21             while (temp != null && temp.left != null) {
22                 temp = temp.left;
23             }
24             return temp;
25         }
26         if (pNode.next != null && pNode.next.left == pNode) {
27             return pNode.next;
28         }
29         while (pNode.next != null && pNode.next.right == pNode) {
30             pNode = pNode.next;
31         }
32         if (pNode.next == null) {
33             return null;
34         }
35         return pNode.next;
36         
37     }
38 }

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我的写法，太复杂，而且会出现空指针异常，因为比如 if (pNode.next != null && pNode.next.left == pNode)

需要先判断pNode.next是否为空，才能判断pNode.next.left

答案写法：

 1 链接：https://www.nowcoder.net/questionTerminal/9023a0c988684a53960365b889ceaf5e
 2 来源：牛客网
 3 
 4 public class Solution {
 5     TreeLinkNode GetNext(TreeLinkNode node)
 6     {
 7         if(node==null) return null;
 8         if(node.right!=null){    //如果有右子树，则找右子树的最左节点
 9             node = node.right;
10             while(node.left!=null) node = node.left;
11             return node;
12         }
13         while(node.next!=null){ //没右子树，则找第一个当前节点是父节点左孩子的节点
14             if(node.next.left==node) return node.next;
15             node = node.next;
16         }
17         return null;   //退到了根节点仍没找到，则返回null
18     }
19 }

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面试题9：用两个栈实现队列

思路：不需要每次都倒腾过去。

push进stack1，pop可以pop到stack2中，从stack2取。

但是只要stack2不空，就可以一直从stack2取，push进去放入stack1。

2018-2-6

面试题10：斐波那契数列

①递归做法：n的指数上升。

②动态规划：T: O(n)　　S:O(1)

f[0] = 0　　f[1] = 1　　f[n] = f[n-1]+f[n-2]

 1 public class Solution {
 2     public int Fibonacci(int n) {
 3         if (n <= 1) {
 4             return n;
 5         }
 6         int[] f = new int[n];
 7         f[0] = 0;
 8         f[1] = 1;
 9         for (int i = 2; i < n; i++) {
10             f[i] = f[i - 1] + f[i - 2];
11         }
12         return f[n - 1] + f[n - 2];
13     }
14 }

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虽然是用动态规划做，但是不需要新建那么大的数组空间int[] f = new int[n];仅需要fn-2 fn-1两个变量即可。

 1 public class Solution {
 2     public int Fibonacci(int n) {
 3         if (n < 2) {
 4             return n;
 5         }
 6         int fnMinus1 = 1;
 7         int fnMinus2 = 0;
 8         int fn = 1;
 9         for (int i = 2; i <= n; i++) {
10             fn = fnMinus1 + fnMinus2;
11             fnMinus2 = fnMinus1; //先改fnMinus2！
12             fnMinus1 = fn;
13         }
14         return fn;
15     }
16 }

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③数学公式　　T：O(logn)

2018-2-8

面试题11：旋转数组的最小数字

面试题12：矩阵中的路径

答案的dfs做法

 1 public class Solution {
 2     public boolean hasPath(char[] matrix, int rows, int cols, char[] str)
 3     {
 4         if(matrix == null || matrix.length < rows * cols || str.length > matrix.length) {
 5             return false;
 6         }
 7         boolean visited[] = new boolean[rows * cols];
 8         //初始化
 9         for (int i = 0; i < rows * cols; i++) {
10             visited[i] = false;
11         }
12         for (int i = 0; i < rows; i++) {
13             for (int j = 0; j < cols; j++) {
14                 if (hasPoint(matrix, visited, rows, cols, i, j, str, 0)){
15                     return true;
16                 }
17             }
18         }
19         return false;
20     }
21     private boolean hasPoint(char[] matrix, boolean[] visited, int rows, int cols, int r, int c, char[] str, int index) {
22 
23         if ((r >= 0 )&& (c >= 0) &&( r < rows) && (c < cols) &&
24             (visited[r * cols + c] == false) && 
25             (matrix[r * cols + c] == str[index])) {
26             if (index == (str.length - 1)) {
27                 return true;
28             }
29             visited[r * cols + c] = true;
30             
31             boolean hasPath = false;
32             index++;
33             hasPath = hasPoint(matrix, visited, rows, cols, r + 1, c, str, index) ||
34                       hasPoint(matrix, visited, rows, cols, r - 1, c, str, index) ||
35                       hasPoint(matrix, visited, rows, cols, r, c + 1, str, index) ||
36                       hasPoint(matrix, visited, rows, cols, r, c - 1, str, index);
37             if (hasPath) {
38                 return true;
39             } else {
40                 visited[r * cols + c] = false;
41             }
42         }
43         return false;
44     }
45 
46 
47 }

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面试题13：机器人的运动范围

我的bfs做法

 1 import java.util.*;
 2 public class Solution {
 3     class Point{
 4         int x;
 5         int y;
 6         public Point(int a, int b) {
 7             x = a;
 8             y = b;
 9         }
10     }
11     public int movingCount(int threshold, int rows, int cols)
12     {
13         int count = 0;
14         if (rows <= 0 || cols <= 0) {
15             return 0;
16         }
17         boolean visited[][] = new boolean[rows][cols];//!!!忘记，默认初始false
18         int[] deltaX ={0, 0, 1, -1};
19         int[] deltaY ={1, -1, 0, 0};
20         Queue<Point> queue = new LinkedList<>();
21         Point p = new Point(0, 0);
22         if (threshold >= 0) {
23             queue.add(p);
24             visited[0][0] = true;//!!!忘记，时刻记着queue.add和visited一起做出改动
25         }
26             
27         
28         while (!queue.isEmpty()) {
29             Point temp = queue.poll();//not pop
30             count++;
31             for (int i = 0; i < 4; i++) {
32                 Point nextPoint = new Point(temp.x + deltaX[i], temp.y + deltaY[i]);
33                 if (isValid(nextPoint, rows, cols, threshold, visited)) {
34                     queue.add(nextPoint);
35                     visited[nextPoint.x][nextPoint.y] = true;
36                     
37                 }
38             }
39         }
40         return count;
41         
42         
43     }
44     private boolean isValid(Point point, int rows, int cols, int threshold, boolean[][] visited) {
45         if (point.x >= 0 && point.x < rows 
46             && point.y >=0 && point.y < cols 
47             && (visited[point.x][point.y] == false)
48             && (countSum(point) <= threshold)) {
49             return true;
50         }else {
51             return false;
52         }
53     }
54     private int countSum(Point p) {
55         int count = 0;
56         int x = p.x;
57         int y = p.y;
58         while (x != 0) {
59             count += x % 10;
60             x = x / 10;
61         }
62         while (y != 0) {
63             count += y % 10;
64             y = y / 10;
65         }
66         return count;
67     }
68 }

View Code

回溯法（答案dfs）

 1 public class Solution {
 2     public int movingCount(int threshold, int rows, int cols)
 3     {
 4         if (threshold < 0 || rows < 0 || cols < 0) {
 5             return 0;
 6         }
 7         boolean[][] visited = new boolean[rows][cols];
 8         return dfs(threshold, rows, cols, 0, 0, visited);
 9         
10     }
11     private int dfs(int threshold, int rows, int cols, int r, int c, boolean[][] visited) {
12         int sum = 0;
13         if (r >= 0 && c >= 0 && r < rows && c < cols && visited[r][c] == false && (count(r) + count(c)) <= threshold) {
14             visited[r][c] = true;
15             sum = dfs(threshold, rows, cols, r + 1, c, visited) +
16                 dfs(threshold, rows, cols, r - 1, c, visited) +
17                 dfs(threshold, rows, cols, r , c + 1, visited) +
18                 dfs(threshold, rows, cols, r , c - 1, visited) + 1;
19         }
20         return sum;
21     }
22     private int count(int x) {
23         int c = 0;
24         while (x != 0) {
25             c += x % 10;
26             x = x / 10;
27         }
28         return c;
29     }
30 }

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面试题15：二进制中1的个数

 1 public class Solution {
 2     public int NumberOf1(int n) {
 3         int count = 0;
 4         while (n != 0) {
 5             count++;
 6             n = (n - 1) & n;
 7         }
 8         return count;
 9     }
10 }

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面试题16：数值的整数次方->时间复杂度O(logn)

　　思路：

　　①判断边界，判断底数base是否为0，double类型不可直接==.

　　　　当底数为0且指数<0时

　　　　会出现对0求倒数的情况，需进行错误处理，设置一个全局变量

　　　　　　0的0次幂和0的负次幂在数学是都是没有定义的

　　②幂正负分情况

　　③快速幂

　　　　　　aⁿ = a^n/2* a^n/2n是偶数

　　　　　　aⁿ = a^n/2* a^n/2 * a n是奇数

　　④exponent >>1 除2写法

　　⑤判断奇偶：(exponent &0x1 == 1)

 1 public class Solution {
 2     public double Power(double base, int exponent) {
 3         if (equal(base, 0.0)&&exponent < 0) {
 4             return 0.0;//无意义，可讨论
 5         }
 6         if (exponent == 0) {
 7             return 1;
 8         }
 9         int n = Math.abs(exponent);
10         double res = Power(base, n >> 1);
11         if ((n & 0x1) == 1) {
12             if (exponent > 0) {
13                 return res * res * base;
14             } else {
15                 return 1 / (res * res * base);
16             }
17         } else {
18             if (exponent > 0) {
19                 return res * res;
20             } else {
21                 return 1 / (res * res);
22             }
23         }
24 
25   }
26     private boolean equal(double a, double b) {
27         if (Math.abs(a - b) < 0.001) {
28             return true;
29         }
30         return false;
31     }
32 }

View Code

稍显复杂

 1 public class Solution {
 2     public double Power(double base, int exponent) {
 3         if (equal(base, 0.0)&&exponent < 0) {
 4             return 0.0;//无意义，可讨论
 5         }
 6         if (exponent < 0) {
 7             exponent = exponent * (-1);
 8             return 1 / help(base, exponent);
 9         }
10         return help(base, exponent);
11   }
12     private double help(double base, int exponent) {
13         if (exponent == 1) {
14             return base;
15         }
16         if (exponent == 0) {
17             return 1;
18         }
19         double res = help(base, exponent >> 1);
20         res = res * res;
21         if ((exponent & 0x1) == 1) {
22             return res * base;
23         } else {
24             return res;
25         }
26     }
27     private boolean equal(double a, double b) {
28         if (Math.abs(a - b) < 0.001) {
29             return true;
30         }
31         return false;
32     }
33 }

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分出来写比较好

double浮点数判断相等

①
final double epsilon = 0.001;
if(Math.abs(double_x - 0) < epsilon)
{
    return true;
}
②
if(Double.doubleToLongBits(double_x) == Double.doubleToLongBits(double_y))

面试题18：删除链表的节点

 1 /*
 2  public class ListNode {
 3     int val;
 4     ListNode next = null;
 5 
 6     ListNode(int val) {
 7         this.val = val;
 8     }
 9 }
10 */
11 public class Solution {
12     public ListNode deleteDuplication(ListNode pHead)
13     {
14         if (pHead == null) {
15             return null;
16         } 
17         ListNode dummy = new ListNode(0);
18         dummy.next = pHead;
19         ListNode pre = dummy;
20         ListNode cur = pHead;
21         ListNode next = cur.next;
22         while (next != null) {
23             while (next!= null && next.val == cur.val) {
24                 cur = next;
25                 next = next.next;
26                 pre.next = cur;
27             }
28             if (next != null) {
29                 pre = cur;
30                 cur = next;
31                 next = next.next;
32             } else {
33                 return dummy.next;
34             }
35             
36         }
37         return dummy.next;
38 
39     }
40 }

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错误的解法，懒得看了。重写

面试题21：调整数组顺序使奇数位于偶数前面

 1 public class Solution {
 2     public void reOrderArray(int [] array) {
 3         if (array == null || array.length == 0) {
 4             return;
 5         }
 6         int left = 0;
 7         int right = 0;
 8         while (left < array.length) {
 9             if (array[left] % 2 == 0) {
10                 right = left + 1;
11                 while (right < array.length && array[right] % 2 == 0) {
12                     right++;
13                 }
14                 int temp = array[left];
15                 array[left] = array[right];
16                 array[right] = temp;
17                 left = right;
18             } else {
19                 left++;
20             }
21         }
22     }
23 }

View Code

错误写法。

面试题19：正则表达式匹配

 1 public class Solution {
 2     public boolean match(char[] str, char[] pattern)
 3     {
 4         if(str == null && pattern == null) {
 5             return true;
 6         } 
 7         if (str.length == 0 && pattern.length == 0) {
 8             return true;
 9         }
10         if (pattern.length == 2 && pattern[0] == '.' && pattern[1] == '*') {
11             return true;
12         }
13         if (str.length == 0 && pattern.length == 2 && pattern[1] == '*') {
14             return true;
15         }
16         if (str == null || pattern == null || str.length == 0 || pattern.length == 0) {
17             return false;
18         }
19 
20         int i = 0;
21         int j = 0;
22         while (i < str.length && j < pattern.length) {
23             if (pattern[j] == '.' || (str[i] == pattern[j])) {
24                 i++;
25                 j++;
26             } else if(str[i] != pattern[j] && (j + 1) < pattern.length && pattern[j + 1] == '*') {
27                 //i++; 不用 分析清楚
28                 j = j + 2;
29             } else if (pattern[j] == '*' && str[i] == pattern[j - 1]) {
30                 i++;
31             } else if (pattern[j] == '*' && (j + 1) < pattern.length && str[i] == pattern[j + 1]) {
32                 j++;
33             }
34             // bbbba      .*a*b        ‘.*’是个大麻烦
35             else if (str[i] != pattern[j] && pattern[j] == '.' && pattern[j + 1] == '*') {
36                 
37             } else {
38                 return false;
39             }
40         }
41         if (i == str.length) {
42             if (j == pattern.length || (j < pattern.length && pattern[j] == '*') || (j == (pattern.length - 2) && pattern[j + 1] == '*')) {
43                  return true;
44             }
45            
46         }
47         return false;
48     }
49 }

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乱写的一堆。

2018-2-14

面试题19：表示数值的字符串

面试题20：正则表达式

面试题27：对称的二叉树

 1 /*
 2 public class TreeNode {
 3     int val = 0;
 4     TreeNode left = null;
 5     TreeNode right = null;
 6 
 7     public TreeNode(int val) {
 8         this.val = val;
 9 
10     }
11 
12 }
13 */
14 public class Solution {
15     boolean isSymmetrical(TreeNode pRoot)
16     {
17         if (pRoot == null) {
18             return true;
19         }
20         return isSym(pRoot.left, pRoot.right);
21     }
22     private boolean isSym(TreeNode left, TreeNode right) {
23         if (left == null && right == null) {
24             return true;
25         }
26         if (left == null) {
27             return false;
28         }
29         if (right == null) {
30             return false;
31         }
32         if (left.val == right.val) {
33             return isSym(left.left, right.right) && isSym(left.right, right.left);
34         }
35         return false;
36     }
37 }

View Code

还是有点糊里糊涂。

面试题30：包含main函数的栈

 1 import java.util.Stack;
 2 
 3 public class Solution {
 4 
 5     Stack<Integer> stack = new Stack<>();
 6     Stack<Integer> minStack = new Stack<>();
 7     public void push(int node) {
 8         if (stack.isEmpty()) {
 9             stack.push(node);
10             minStack.push(node);
11         } else {
12             stack.push(node);
13             if (node < minStack.peek()) {
14                 minStack.push(node);
15             } else {
16                 minStack.push(minStack.peek());
17             }
18         }
19     }
20     
21     public void pop() {
22         stack.pop();
23         minStack.pop();
24     }
25     
26     public int top() {
27         return stack.peek();
28     }
29     
30     public int min() {
31         return minStack.peek();
32     }
33 }

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不用每个都加一次

 1 import java.util.Stack;
 2 
 3 public class Solution {
 4 
 5     Stack<Integer> stack = new Stack<>();
 6     Stack<Integer> minStack = new Stack<>();
 7     public void push(int node) {
 8         if (stack.isEmpty()) {
 9             stack.push(node);
10             minStack.push(node);
11         } else {
12             stack.push(node);
13             if (node < minStack.peek()) {
14                 minStack.push(node);
15             } 
16         }
17     }
18     
19     public void pop() {
20         int node = stack.pop();
21         if (node <= minStack.peek()) {
22             if (stack.peek() > minStack.peek()) {
23                 minStack.pop();
24             }
25         }
26     }
27     
28     public int top() {
29         return stack.peek();
30     }
31     
32     public int min() {
33         return minStack.peek();
34     }
35 }

View Code

面试题34：二叉树中和为某一值的路径

 1 import java.util.ArrayList;
 2 /**
 3 public class TreeNode {
 4     int val = 0;
 5     TreeNode left = null;
 6     TreeNode right = null;
 7 
 8     public TreeNode(int val) {
 9         this.val = val;
10 
11     }
12 
13 }
14 */
15 public class Solution {
16     public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target) {
17         ArrayList<ArrayList<Integer>> paths = new ArrayList<>();
18         if (root == null) {
19             return paths;
20         }
21         
22         ArrayList<Integer> path = new ArrayList<>();
23         help(root, target, paths, path);
24         return paths;
25     }
26     private void help(TreeNode root, int target, ArrayList<ArrayList<Integer>> paths, ArrayList<Integer> path) {
27         //这样写有问题，叶子的 左null 右null会放入两次
28         //测试用例:{10,5,12,4,7},22
29         //对应输出应该为:[[10,5,7],[10,12]]
30         //你的输出为:[[10,5,7],[10,5,7],[10,12],[10,12]]
31         //if (target == 0 && root == null) { //从根到叶不能断
32         //    paths.add(new ArrayList<Integer>(path));
33         //    return;
34         //}
35         if (target < 0 || root == null) {
36             return;
37         }
38         if (target == root.val && root.left == null && root.right == null) {
39             path.add(root.val);
40             paths.add(new ArrayList<Integer>(path));
41             path.remove(path.size() - 1);
42             return;
43         }
44         path.add(root.val);
45         help(root.left, target - root.val, paths, path);
46         help(root.right, target - root.val, paths, path);
47         path.remove(path.size() - 1);
48     }
49 }

View Code

面试题35：复杂链表的复制

 1 /*
 2 public class RandomListNode {
 3     int label;
 4     RandomListNode next = null;
 5     RandomListNode random = null;
 6 
 7     RandomListNode(int label) {
 8         this.label = label;
 9     }
10 }
11 */
12 public class Solution {
13     public RandomListNode Clone(RandomListNode pHead)
14     {
15         if (pHead == null) {
16             return null;
17         }
18         RandomListNode cur = pHead;
19         while (cur != null) {
20             RandomListNode temp = new RandomListNode(cur.label);
21             temp.next = cur.next;
22             cur.next = temp;
23             cur = temp.next;
24         }
25         cur = pHead;
26         while (cur != null) {
27             if (cur.random != null) {
28                 cur.next.random = cur.random.next;
29             }
30             cur = cur.next.next;
31         }
32         RandomListNode copyHead = pHead.next;
33         cur = pHead;
34         RandomListNode copyCur = copyHead;
35         while (copyCur.next != null) { //这个判断条件就这么简单，要好好想想
36             cur.next = cur.next.next;
37             copyCur.next = copyCur.next.next;
38             cur = cur.next;
39             copyCur = copyCur.next;
40         }
41         cur.next = null; //最后一个的判断要注意
42         return copyHead;
43     }
44 }

View Code

1->1'->2->2'->null

因为cur = 1, copyCur = 1',

cur = cur.next.next;　　cur.next一定存在

但是copyCur.next可能为null，所以while条件是copyCur.next != null

出来时cur=2， copycur = 2',所以要把cur.next = null

面试题37：序列化二叉树

 1 /*
 2 public class TreeNode {
 3     int val = 0;
 4     TreeNode left = null;
 5     TreeNode right = null;
 6 
 7     public TreeNode(int val) {
 8         this.val = val;
 9 
10     }
11 
12 }
13 */
14 import java.util.*;
15 public class Solution {
16     String Serialize(TreeNode root) {
17         if (root == null) {
18             return "{}";
19         }
20         Queue<TreeNode>  queue = new LinkedList<>();
21         queue.add(root);
22         StringBuffer sb = new StringBuffer();
23         sb.append("{");
24         while (!queue.isEmpty()) {
25             TreeNode temp = queue.poll();
26             if (temp != null) {
27                 sb.append(temp.val);
28                 sb.append(",");
29                 queue.add(temp.left);
30                 queue.add(temp.right);
31             } else {
32                 sb.append("#,");
33             }
34         }
35         sb.deleteCharAt(sb.length() - 1);
36         sb.append("}");
37         return sb.toString();
38   }
39     TreeNode Deserialize(String str) {
40        if (str == null || str.length() == 0 || str.equals("{}")) {
41            return null;
42        }
43         str = str.substring(1, str.length() - 1);
44        String[] array = str.split(",");
45        TreeNode root = new TreeNode(Integer.parseInt(array[0]));
46         Queue<TreeNode> queue = new LinkedList<>();
47         queue.add(root);
48        int i = 1;
49        while (!queue.isEmpty()) {
50            TreeNode temp = queue.poll();
51            if ("#".equals(array[i])) {
52                temp.left = null;
53            } else {
54                TreeNode leftNode = new TreeNode(Integer.parseInt(array[i]));
55                temp.left = leftNode;
56                queue.add(leftNode);
57            }
58            i++;
59            if ("#".equals(array[i])) {
60                temp.right = null;
61            } else {
62                TreeNode rightNode = new TreeNode(Integer.parseInt(array[i]));
63                temp.right = rightNode;
64                queue.add(rightNode);
65            }
66            i++;
67        }
68         return root;
69   }
70 }

View Code

str = str.substring(1, str.length() - 1);
String：[] array = str.split(",");

连续子数组的最大和

 1 public class Solution {
 2     public int FindGreatestSumOfSubArray(int[] array) {
 3         if (array == null || array.length == 0) {
 4             return 0;
 5         }
 6         int max = Integer.MIN_VALUE;
 7         int sum = 0;
 8         for (int i = 0; i < array.length; i++) {
 9             if (sum < 0) {
10                 sum = 0;
11             }
12             sum = sum + array[i];                       
13             if (sum > max) {
14                 max = sum;
15             }
16         }
17         return max;
18     }
19 }

View Code

数组中出现次数超过一半的数字

 1 import java.util.*;
 2 public class Solution {
 3     public int MoreThanHalfNum_Solution(int [] array) {
 4         if (array == null || array.length == 0) {
 5              return 0;
 6         }
 7         if (array.length == 1) {
 8             return array[0];
 9         }
10         Map<Integer, Integer> map = new HashMap<Integer, Integer>();
11         int length = array.length / 2;
12         for (int i = 0; i < array.length; i++) {
13             if (map.containsKey(array[i])) {
14 
15                 map.put(array[i], map.get(array[i]) + 1);
16                 if (map.get(array[i]) > length) {
17                     return array[i];
18                 }
19             } else {
20                 map.put(array[i], 1);
21             }
22         }
23         return 0;
24     }
25 }

View Code

最小的k个数

 1 import java.util.*;
 2 public class Solution {
 3     public ArrayList<Integer> GetLeastNumbers_Solution(int [] input, int k) {
 4         ArrayList<Integer> res = new ArrayList<>();
 5         if (input == null || input.length < k) {
 6             return res;
 7         }
 8         PriorityQueue<Integer> minPq = new PriorityQueue<>();
 9         for (int i = 0; i < input.length; i++) {
10             minPq.offer(input[i]);
11         }
12         for (int i = 0; i < k; i++){
13             res.add(minPq.poll());
14         }
15         return res;
16     }
17 }

View Code

答案

 1 public class Solution {
 2     public int MoreThanHalfNum_Solution(int [] array) {
 3         if (array == null || array.length == 0) {
 4             return 0;
 5         }
 6         int res = array[0];
 7         int count = 1;
 8         for (int i = 1; i < array.length; i++) {
 9             if (array[i] != array[i - 1]) {
10                 count--;
11             } else {
12                 count++;
13             }
14             if (count == 0) {
15                 count  = 1;
16                 res = array[i];
17             }
18         }
19             //check
20         int doubleCount = 0;
21         for (int i = 0; i < array.length; i++) {
22             if (array[i] == res) {
23                 doubleCount++;
24             }
25          }
26          if (doubleCount > array.length / 2) {
27              return res;
28          } 
29          return 0;
30     }
31 }

View Code

转载于:https://www.cnblogs.com/yunyouhua/p/8418700.html

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原文地址：https://www.cnblogs.com/twodog/p/12137555.html