数组

Tips:

1.遇到subArray相关问题，多思考prefix sum

2.2Sum相关问题的两种解法

　　hashMap:更加适用于求解结果和下标相关的问题

　　sortedArray + twoPoint：更加适用于求解结果和值相关问题。对于closest问题多半是使用 two point

3.Partition Array 二分三分

Merge Two Sorted Arrays

 1 public int[] mergeSortedArray(int[] A, int[] B) {
 2         // Write your code here
 3         if (A == null) {
 4             return B;
 5         }
 6         if (B == null) {
 7             return A;
 8         }
 9         int m = A.length;
10         int n = B.length;
11         int[] results = new int[m + n];
12         int i = 0;
13         int j = 0;
14         int k = 0;
15         while (i < m && j < n) {
16             if (A[i] < B[j]) {
17                 results[k++] = A[i++];
18             } else {
19                 results[k++] = B[j++];
20             }
21         }
22         while (i < m) {
23             results[k++] = A[i++];
24         }
25         while (j < n) {
26             results[k++] = B[j++];
27         }
28         return results;
29     }

View Code

Merge Sorted Array

Given two sorted integer arrays A and B, merge B into A as one sorted array.

 1 public void mergeSortedArray(int[] A, int m, int[] B, int n) {
 2         // write your code here
 3         int k = m + n - 1;
 4         while (m > 0 && n > 0) {
 5             if (A[m - 1] > B[n - 1]) {
 6                 A[k--] = A[--m];
 7             } else {
 8                 A[k--] = B[--n];
 9             }
10         }
11         while (n > 0) {
12             A[k--] = B[--n];
13         }
14     }

View Code

Intersection of Two Arrays

Given two arrays, write a function to compute their intersection.

使用两个hash避免使用一个hash的删除操作

 1 public int[] intersection(int[] nums1, int[] nums2) {
 2         // Write your code here
 3         if (nums1 == null || nums2 == null) {
 4             return null;
 5         }
 6         HashSet<Integer> hashSet = new HashSet<Integer>();
 7         HashSet<Integer> resultHash = new HashSet<Integer>();
 8         for (int i = 0; i < nums1.length; i++) {
 9             hashSet.add(nums1[i]);
10         }
11         for (int i = 0; i < nums2.length; i++) {
12             if (hashSet.contains(nums2[i]) && !resultHash.contains(nums2[i])) {
13                 resultHash.add(nums2[i]);
14             }
15         }
16         int[] results = new int[resultHash.size()];
17         int index = 0;
18         for (int i : resultHash) {
19             results[index++] = i;
20         }
21         return results;
22     }

View Code

排序之后再操作--not bug free

 1  public int[] intersection(int[] nums1, int[] nums2) {
 2         // Write your code here
 3         if (nums1 == null || nums2 == null) {
 4             return null;
 5         }
 6         Arrays.sort(nums1);
 7         Arrays.sort(nums2);
 8         int[] results = new int[nums1.length];
 9         int i = 0;
10         int j = 0;
11         int index = 0;
12         while (i < nums1.length && j < nums2.length) {
13             if (nums1[i] == nums2[j]) {
14                 if (index == 0 || results[index - 1] != nums1[i]) {
15                     results[index++] = nums1[i];
16                 }
17                 i++;
18                 j++;
19             } else if (nums1[i] < nums2[j]) {
20                 i++;
21             } else {
22                 j++;
23             }
24         }
25         int[] res = new int[index];
26         for (int k = 0; k < index; k++) {
27             res[k] = results[k];
28         }
29         return res;
30     }

View Code

Median of two Sorted Arrays--not bug free

There are two sorted arrays A and B of size m and nrespectively. Find the median of the two sorted arrays.

 1 public double findMedianSortedArrays(int[] A, int[] B) {
 2         // write your code here
 3         if (A == null && B == null) {
 4             return 0;
 5         } 
 6         if (A == null) {
 7             if (B.length % 2 == 0) {
 8                 return (B[B.length / 2 - 1] + B[B.length / 2]) / 2.0;
 9             }
10             else {
11                 return B[B.length / 2];
12             }
13         } 
14         if (B == null) {
15             if (A.length % 2 == 0) {
16                 return (A[A.length / 2 - 1] + A[A.length / 2]) / 2.0;
17             }
18             else {
19                 return A[A.length / 2];
20             }
21         }
22         int m = A.length;
23         int n = B.length;
24         if ((m + n) % 2 == 0) {
25             return (findTopK(A, B, 0, 0, (m + n) / 2) 
26                 + findTopK(A, B, 0, 0, (m + n) / 2 + 1)) / 2.0;
27         } else {
28             return findTopK(A, B, 0, 0, (m + n) / 2 + 1);
29         }
30     }
31     public int findTopK(int[] A, int[] B, int AStart, int BStart,int topK) {
32         if (AStart > A.length  - 1) {
33             return B[BStart + topK - 1];
34         }
35         if (BStart > B.length - 1) {
36             return A[AStart + topK - 1];
37         }
38         if (topK == 1) {
39             return Math.min(A[AStart], B[BStart]);
40         }
41         if (AStart + topK / 2 - 1 > A.length - 1) {
42             return findTopK(A, B, AStart, BStart + topK / 2, topK - topK / 2);
43         }
44         if (BStart + topK / 2 - 1 > B.length - 1) {
45             return findTopK(A, B, AStart + topK / 2, BStart, topK - topK / 2);
46         }
47         if (A[AStart + topK / 2 - 1] < B[BStart + topK / 2 - 1]) {
48             return findTopK(A, B, AStart + topK / 2, BStart, topK - topK / 2);
49         } else {
50             return findTopK(A, B, AStart, BStart + topK / 2, topK - topK / 2);
51         }
52     }

View Code