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

The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 has a size equal to m + n such that it has enough space to hold additional elements from nums2.

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]

Constraints:

nums1.length == m + n
nums2.length == n
0 <= m, n <= 200
1 <= m + n <= 200
-109 <= nums1[i], nums2[i] <= 109

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

思路很好想~ 两个数组nums1和nums2都是已经排好序的，那么我们只需要将排好序的数组进行逐个比较数值即可

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int i=0,j=0;
        vector<int> nums3;
        for(;i<m&&j<n; ){
           
            if(nums1[i]<nums2[j]){
                
                nums3.push_back(nums1[i]);i++;
                
            }else if(nums1[i] == nums2[j]){
                nums3.push_back(nums1[i]);
                i++;
                nums3.push_back(nums2[j]);
                j++;
                
            }else {
                nums3.push_back(nums2[j]);
                j++;
            }
        }
        while(i<m) nums3.push_back(nums1[i++]);
        while(j<n) nums3.push_back(nums2[j++]);
        for(j=0; j<size(nums3); j++){
            
            nums1[j] = nums3[j];
        }
    }
};