My first reaction: move all A elements back by n positions, and start everything from A[0] and B[0]. But a smarter idea is to start everything from the end :) So no need to move. Just to take care of all cases.
class Solution { public: void merge(int A[], int m, int B[], int n) { int cnt = m + n - 1; int bi = n - 1, ai = m - 1; while (cnt >= 0) { int tmp = 0; if (ai < 0 && bi >= 0) { tmp = B[bi--]; } else if (bi < 0 && ai >= 0) { tmp = A[ai--]; } else if (bi >=0 && ai >= 0) { if (A[ai] >= B[bi]) tmp = A[ai--]; else tmp = B[bi--]; } A[cnt--] = tmp; } } };
LeetCode "Merge Sorted Array",布布扣,bubuko.com
原文：http://www.cnblogs.com/tonix/p/3857891.html