Merge Sorted Array解题报告

Description:

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.

Link:

https://leetcode.com/problems/merge-sorted-array/#/description

解题方法：

先把nums1的元素整体往后挪n位，然后在以nums1为容器排序。

Time Complexity：

空间：O(1)
时间：O(N)

完整代码：

void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) 
    {
        for(int i = m - 1, j = m + n - 1; i >= 0; i--, j--)
            nums1[j] = nums1[i];
        int start1 = n;
        int start2 = 0;
        int move = 0;
        while(start1 < m + n || start2 < n)
        {
            if(start2 >= n)
            {
                nums1[move++] = nums1[start1++];
                continue;
            }
            if(start1 >= m + n)
            {
                nums1[move++] = nums2[start2++];
                continue;
            }
            if(nums1[start1] < nums2[start2])
                nums1[move++] = nums1[start1++];
            else
                nums1[move++] = nums2[start2++];
        }
    }