88 - Merge Sorted Array

#easy

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.

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

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        if ( n == 0 ) return;
        int i = m - 1;  // i for nums1
        int j = n - 1;   // j for nums2
        for ( int k = nums1.size() - 1; k >= 0; k-- ) { // k for iterator
            if ( j < 0 || ( i >= 0 && nums1[i] > nums2[j] ) ) { // key point to decide i, j, k
                nums1[k] = nums1[i];
                i--;
            }
            else {
                nums1[k] = nums2[j];
                j--;
            }
        }
    }
};