[LeetCode] Merge Sorted Array

Problem Statement (link):

Given two sorted integer arrays A and B, merge B into A as one sorted array.
Note:
You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m andn respectively.

Analysis:
The obvious way is to insert B's entries into A after each comparison, and shift all the rest of A to right. Repeating this until we reach the end of B. I bet you'll get TLE.

The way to improve is considering about avoid these shifts. As the A[m+1 : m+n] is empty, we could start from the large values and works backwards. See the following implementation. The time complexity is O(m+n) and no extra space is needed.

Code:

class Solution {
public:
    void merge(int A[], int m, int B[], int n) {
        if (n==0) return;
        if (m==0) {
            for (int i=0; i<n; i++)
                A[i]=B[i];
            return;
        }

        int curA=m; int curB=n;
        while(curB>0 && curA>0) {
            if (A[curA-1]>=B[curB-1]) {
                A[curA+curB-1]=A[curA-1];
                curA--;
            }
            else  {
                A[curA+curB-1]=B[curB-1];
                curB--;
            }
        }

        if (curB>0)
            for (int i=0; i<curB; i++)
                A[i]=B[i];
        if (curA>0)
            return;
    }
};