剑指offer(1-10)

题目一

在一个二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。

C++代码如下（暴力解法）：

class Solution {
public:
    bool Find(int target, vector<vector<int> > array) {
        bool isFound = false;
        for(int i = 0; i < array.size(); i++){
            for(int j = 0; j < array[i].size(); j++){
                if(false == isFound && target == array[i][j]){
                    isFound = true;
                }
            }
        }
        return isFound;
    }
};

JAVA代码如下（暴力解法）：

public class Solution {
    public boolean Find(int target, int [][] array) {
        boolean Flag = false;
        for(int i = 0; i < array.length; i++){
            for(int j = 0; j < array[i].length; j++) {
                if(Flag == false && array[i][j] == target)
                    Flag = true;
            }
        }
        return Flag;
    }

C++最优解如下：

class Solution {
public:
    bool Find(int target, vector<vector<int> > array) {
        bool isFound = false;
        int row = 0;
    	int col = array[0].size() - 1;
    	while (row < array.size() && col >= 0) {
            if(target > array[row][col]) {
                row++;
            }else if(target < array[row][col]) {
                col--;
            }else {
                isFound = true;
                break;
            }
		}
        return isFound;
    }
};

JAVA最优解如下：

public class Solution {
    public boolean Find(int target, int [][] array) {
        boolean Flag = false;
    	int row = 0;
    	int col = array[0].length - 1;
    	while (row < array.length && col >= 0) {
            if(target > array[row][col]) {
                row++;
            }else if(target < array[row][col]) {
                col--;
            }else {
                Flag = true;
                break;
            }
		}
    	return Flag;
    }
}

题目二

请实现一个函数，将一个字符串中的空格替换成“%20”。例如，当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy。
C++暴力解如下：

class Solution {
public:
	void replaceSpace(char *str,int length) {
        if (str == NULL || length <= 0)
			return;
	int i = length - 1, j;
	int len = length;
	//从后往前进行遍历 

	for (int i = length - 1; i >= 0; i--) {
		if (str[i] == ' ') {
		len += 2;
		for (j = len - 1; j > i+2; j--) {
			str[j] = str[j - 2];
		}
		str[i] = '%';
		str[i + 1] = '2';
		str[i + 2] = '0';
		}
	}
	str[len] = ''; 
	}
};

C++最优解如下：

class Solution {
public:
	void replaceSpace(char *str,int length) {
        if (str == nullptr && length < 0)
			return;
	//获得原始字符串长度和空格数   

	int OrignalLength = 0;
	int NumberOfBlank = 0;
	int i = 0;
	while (str[i] != '') {
		OrignalLength++;
		if (str[i] == ' ')
			NumberOfBlank++;
		++i;
	}
	//将字符串空格替换为%20   

	int NewLength = OrignalLength + 2 * NumberOfBlank;
	if (NewLength > length)
		return;
	int indexOfOriginal = OrignalLength;
	int indexOfNew = NewLength;
	while (indexOfOriginal >= 0 && indexOfNew > indexOfOriginal) {
		if (str[indexOfOriginal] == ' ') {
			str[indexOfNew--] = '0';
			str[indexOfNew--] = '2';
			str[indexOfNew--] = '%';
			indexOfOriginal--;
		}else { 
			str[indexOfNew--] = str[indexOfOriginal--];
		}	
	}
    }
};

JAVA最优解如下：

public class Solution {
    public String replaceSpace(StringBuffer str) {
    	if(str == null)
		return null;
	StringBuilder sb = new StringBuilder();
	for(int i = 0; i < str.length(); i++) {
		if(String.valueOf(str.charAt(i)).equals(" ")) {
			sb.append("%20");
		}else {
			sb.append(str.charAt(i));
		}
	}
	return String.valueOf(sb);
    }
}

题目三

输入一个链表，从尾到头打印链表每个节点的值。

C++最优解如下：

//利用栈和容器逆向打印链表   

vector<int> printListFromTailToHead04(ListNode* head) {
    vector <int> nodeVector;
    stack<int> stack;
    ListNode *p = head;
    if (head != nullptr) {
        while (p != nullptr) {
            stack.push(p->val);
            p = p->next;
        }
        while (!stack.empty()) {
            nodeVector.push_back(stack.top());
            stack.pop();
        }
    }
    return nodeVector;

}

JAVA最优解如下：

import java.util.ArrayList;
import java.util.Stack;
public class Solution {
    public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
        Stack<Integer> stack = new Stack<Integer>(); 
	    while(listNode != null) {
	    	stack.push(listNode.val);
	    	listNode = listNode.next;
	    }
	    
	    ArrayList<Integer> list = new ArrayList<Integer>();
	    while(!stack.empty()) {
	    	list.add(stack.pop());
	    }
		return list;
    }
}

题目四

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

C++最优解如下：

TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> vin) {
        int inLength = vin.size();
	if (inLength == 0)
		return NULL;

	vector<int> left_pre, right_pre, left_in, right_in;

	//创建根节点，根节点是前序遍历的第一个数  

	TreeNode* head = new TreeNode(pre[0]);

	//找到中序遍历根节点所在位置，存放在变量rootIndex中  

	int rootIndex = 0;
	for (int i = 0; i < inLength; i++) {
		if (vin[i] == pre[0]) {
			rootIndex = i;
			break;
		}
	}

	//对于中序遍历，根节点左边的节点位于二叉树的左边，根节点右节点位于二叉树的右边  

    //利用上述这点，对二叉树进行归并   

	for (int i = 0; i < rootIndex; i++) {
		left_in.push_back(vin[i]);
		left_pre.push_back(pre[i+1]);//前序第一个为根节点
	}
	for (int i = rootIndex + 1; i < inLength; i++) {
		right_in.push_back(vin[i]);
		right_pre.push_back(pre[i]);
	}

	//取出前序和中序遍历根节点左子树和右字树，进行递归再区分左右字树直到叶节点   

	head->left = reConstructBinaryTree(left_pre, left_in);
	head->right = reConstructBinaryTree(right_pre, right_in);
	return head;
}

JAVA最优解如下：

	//已知前序和中序遍历，重建二叉树返回头结点    
    
	public static TreeNode reConstructBinaryTree(int [] pre,int [] in) {
		TreeNode root = reConstructBinaryTree(pre, 0, pre.length-1, in, 0, in.length-1);
		return root;
        }
	
	private static TreeNode reConstructBinaryTree(int[] pre,int startPre,int endPre,int[] in,int startIn,int endIn) {
		if(startPre > endPre || startIn > endIn)
			return null;
		TreeNode root = new TreeNode(pre[startPre]);
		
		for(int i = startIn; i <= endIn; i++) {
			if(in[i] == pre[startPre]) {
				root.left = reConstructBinaryTree(pre, startPre+1, startPre+i-startIn, in, startIn, i-1);
				root.right = reConstructBinaryTree(pre, i-startIn+startPre+1, endPre, in, i+1, endIn);
				break;
			}
		}
		return root;
	}

题目五

用两个栈来实现一个队列，完成队列的Push和Pop操作。 队列中的元素为int类型。

C++最优解如下：

class Solution
{
public:
	void push(int node) {
		stack1.push(node);
	}
	int pop() {
	    if (stack2.empty()) {
		while (!stack1.empty()) {
		    int value = stack1.top();
		    stack2.push(value);
	            stack1.pop();
		}
	    }
	    int value = stack2.top();
	    stack2.pop();
	    return value;
	}
private:
	stack<int> stack1;
	stack<int> stack2;
};

JAVA最优解如下：

import java.util.Stack;

public class Solution {
    Stack<Integer> stack1 = new Stack<Integer>();
    Stack<Integer> stack2 = new Stack<Integer>();
    
    public void push(int node) {
        stack1.push(node);
    }
    
    public int pop() {
        if(stack2.empty()) {
            while(!stack1.empty()) {
                stack2.push(stack1.pop());
            }
        }
        return stack2.pop();
    }
}

题目六

把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。 输入一个非递减排序的数组的一个旋转，输出旋转数组的最小元素。 例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1。 NOTE：给出的所有元素都大于0，若数组大小为0，请返回0。

C++暴力解如下：

class Solution {
public:
    int minNumberInRotateArray(vector<int> rotateArray) {
        int length = rotateArray.size();
	if (length == 0)
		return 0;
	for (int i = 0; i <= length-1; i++) {
		if (rotateArray[i + 1] < rotateArray[i]) {
			return rotateArray[i + 1];
		}
	}
	return rotateArray[0];
    }
};

JAVA暴力解如下：

import java.util.ArrayList;
public class Solution {
   public int minNumberInRotateArray(int[] array) {
        if (array.length == 0)
            return 0;
        for (int i = 0; i < array.length - 1; i++) {
            if (array[i] > array[i + 1])
                return array[i + 1];
        }
        return array[0];
    } 
}

C++最优解如下：

class Solution {
public:
    int minNumberInRotateArray(vector<int> rotateArray) {
        int low = 0 ; int high = rotateArray.size() - 1;   
        while(low < high){
            int mid = low + (high - low) / 2;        
            if(rotateArray[mid] > rotateArray[high]){
                low = mid + 1;
            }else if(rotateArray[mid] == rotateArray[high]){
                high = high - 1;
            }else{
                high = mid;
            }   
        }
        return rotateArray[low];
    }
};

JAVA最优解如下：

import java.util.ArrayList;
public class Solution {
   public int minNumberInRotateArray(int[] array) {
       if (array.length == 0)
           return 0;
       int left = 0;
       int right = array.length - 1;
       int middle = -1;
       while (array[left]>=array[right]) {
           if(right-left==1){
               middle = right;
               break;
           }
           middle = left + (right - left) / 2;
           if (array[middle] >= array[left]) {
               left = middle;
           }
           if (array[middle] <= array[right]) {
               right = middle;
           }
       }
       return array[middle];
    } 
}

题目七

大家都知道斐波那契数列，现在要求输入一个整数n，请你输出斐波那契数列的第n项。n<=39

C++最优解如下：

class Solution {
public:
    int Fibonacci(int n) {
	int result[2] = {0,1};
	if (n < 2) {
		return result[n];
	}
	long long fibNMinusOne = 1;
	long long fibNMinusTwo = 0;
	long long fibN = 0;
	for (int i = 2; i <= n; i++) {
		fibN = fibNMinusOne + fibNMinusTwo;
		fibNMinusTwo = fibNMinusOne;
		fibNMinusOne = fibN;
	}
	return fibN;
    }
};

JAVA最优解如下：

public class Solution {
    public int Fibonacci(int n) {
        int[] result = {0,1};
	if(n < 2) {
		return result[n];
	}
	int FiboNMinusOne = 1;
	int FiboNMinusTwo = 0;
	int FiboN = 0;
	for(int i=2; i <= n; i++) {
		FiboN = FiboNMinusOne + FiboNMinusTwo;

		FiboNMinusTwo = FiboNMinusOne;
		FiboNMinusOne = FiboN;
	}
	return FiboN;
    }
}

题目八

一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法。

C++最优解如下：

class Solution {
public:
    int jumpFloor(int number) {
        int result[2] = {1,2};
	if(number <= 2)
		return result[number-1];
	int FiboNMinusOne = 2;
	int FiboNMinusTwo = 1;
	int FiboN = 0;
	for(int i=3; i <= number; i++) {
		FiboN = FiboNMinusOne + FiboNMinusTwo;
		FiboNMinusTwo = FiboNMinusOne;
		FiboNMinusOne = FiboN;
	}
	return FiboN;
    }
};

JAVA最优解如下：

public class Solution {
    public int JumpFloor(int target) {
        int[] result = {1,2};
	if(target <= 2)
		return result[target-1];
	int FiboNMinusOne = 2;
	int FiboNMinusTwo = 1;
	int FiboN = 0;
	for(int i=3; i <= target; i++) {
		FiboN = FiboNMinusOne + FiboNMinusTwo;
		FiboNMinusTwo = FiboNMinusOne;
		FiboNMinusOne = FiboN;
	}
	return FiboN;
    }
}

题目九

一只青蛙一次可以跳上1级台阶，也可以跳上2级……它也可以跳上n级。求该青蛙跳上一个n级的台阶总共有多少种跳法。

C++最优解如下：

class Solution {
public:
    int jumpFloorII(int number) {
        int result[2] = {0,1};
        if(number == 1 || number == 0)
            return result[number];
        int initialNum = 1;
        int jumpNum = 0;
        for(int i=2; i <= number; i++){
            jumpNum = 2 * initialNum;
            initialNum = jumpNum;
        }
        return jumpNum;
    }
};

JAVA最优解如下：

public class Solution {
    public int JumpFloorII(int target) {
        int[] result = {0,1};
        if(target == 1 || target == 0)
            return result[target];
        int initialNum = 1;
        int jumpNum = 0;
        for(int i=2; i <= target; i++){
            jumpNum = 2 * initialNum;
            initialNum = jumpNum;
        }
        return jumpNum;
    }
}

题目十

我们可以用2*1的小矩形横着或者竖着去覆盖更大的矩形。请问用n个2*1的小矩形无重叠地覆盖一个2*n的大矩形，总共有多少种方法？

C++最优解如下：

class Solution {
public:
    int rectCover(int number) {
        int result[2] = {1,2};
        if(number == 1 || number == 2)
            return result[number-1];
        int fiboNMinusOne = 2;
        int fiboNMinusTwo = 1;
        int fibN = 0;
        for(int i=3; i<=number; i++){
            fibN = fiboNMinusOne + fiboNMinusTwo;
            fiboNMinusTwo = fiboNMinusOne;
            fiboNMinusOne = fibN;
        }
        return fibN;
    }
};

JAVA最优解如下：

public class Solution {
    public int RectCover(int target) {
        int[] result = {1,2};
        if(target == 1 || target == 2)
            return result[target-1];
        int fiboNMinusOne = 2;
        int fiboNMinusTwo = 1;
        int fibN = 0;
        for(int i=3; i<=target; i++){
            fibN = fiboNMinusOne + fiboNMinusTwo;
            fiboNMinusTwo = fiboNMinusOne;
            fiboNMinusOne = fibN;
        }
        return fibN;
    }
}