You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.



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]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.



Constraints:

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

  • nums1的太小为n +m, 那么从最后面开始填充数并不会影响nums1前面的数,用2个指针分别指向nums1和nums2的末尾，进行比较，较大的数填充到nums1的尾部。
class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int idx = m + n - 1;
        m--, n--;
        while (m >= 0 && n >= 0) {
            if (nums1[m] > nums2[n]) nums1[idx--] = nums1[m--];
            else nums1[idx--] = nums2[n--];
        }        
        while (n >= 0) nums1[idx--] = nums2[n--];
    }
};