88. Merge Sorted Array【easy】

88. Merge Sorted Array【easy】

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.

解法一：

 1 class Solution {
 2 public:
 3     void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
 4         int i = m - 1;
 5         int j = n - 1;
 6         int len = m + n - 1;
 7         
 8         if (n == 0)
 9         {
10             return;
11         }
12         
13         while (i >=0 && j >=0)
14         {
15             if (nums1[i] >= nums2[j])
16             {
17                 nums1[len--] = nums1[i--];
18             }
19             else
20             {
21                 nums1[len--] = nums2[j--];
22             }
23         }
24         
25         /*
26         while (i >= 0)
27         {
28             nums1[len--] = nums1[i--];
29         }
30         */
31         
32         while (j >= 0)
33         {
34             nums1[len--] = nums2[j--];
35         }
36     }
37 };

从后向前搞，nums2完毕之后就不用处理nums1了，因为都是往nums1中合并的

解法二：

1 class Solution {
2 public:
3     void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
4         int i = m - 1, j = n - 1, tar = m + n - 1;
5         while (j >= 0) {
6             nums1[tar--] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
7         }
8     }
9 };

This code relies on the simple observation that once all of the numbers from nums2 have been merged into nums1, the rest of the numbers in nums1 that were not moved are already in the correct place.