LeetCode——Merge Sorted Array-优快云博客

本文介绍了一种高效的数组合并方法，通过从后向前比较两个已排序数组的元素，并将较大的元素放置到目标数组的末尾，实现了O(n)的时间复杂度。

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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 to hold additional elements from B. The number of elements initialized in A and B are m and n respectively.

» Solve this problem

假设A = [2, 3, 4, 5] B = [1, 1]

如果顺序合并，

A[0] > B[0] ，将A往后移动一个，B[0]加入 => A = [1, 2, 3, 4, 5]；

A[1] > B[1]，将A往后移动一个，B[1]加入 => A = [1, 1, 2, 3, 4, 5]。

当B[i] < A[j]时，A都要移动整个数组，这种做法复杂度将为O(n^2)。

不妨考虑倒着合并，

A[3] > B[1]，所以A[5] = A[3] => A = [2, 3, 4, 5, x, 5]

A[2] > B[1]，所以A = [2, 3, 4, 5, 4, 5]

...

这种做法复杂度为O(n)。

class Solution {
public:
    void merge(int A[], int m, int B[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int idxA = m - 1, idxB = n - 1;
        for (int i = m + n - 1; i >= 0; i--) {
            if (idxA == -1 || (idxB != -1 && A[idxA] <= B[idxB])) {
                A[i] = B[idxB--];
            }
            else {
                A[i] = A[idxA--];
            }            
        }
    }
};