Leetcode 718. Maximum Length of Repeated Subarray

本题出处

本题是来自于Leetcode的一道题目，算是一道经典题目。最大公共子串问题。此处温馨提醒，注意区分最大公共子串和最大公共子序列。

最大公共子串 VS 最大公共子序列

问题描述：

Given two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.

---

Example 1:
Input: A: [1,2,3,2,1] B: [3,2,1,4,7]
Output: 3

Explanation:
The repeated subarray with maximum length is [3, 2, 1].
Note: 1 <= len(A), len(B) <= 1000
0 <= A[i], B[i] < 100

Solution:

class Solution {public:    int findLength(vector<int>& A, vector<int>& B) {        int ret = 0;        int am[1000][1000] = {0};        for (int x = 0; x < A.size(); ++x) {            for (int y = 0; y < B.size(); ++y)                if (A[x] == B[y]) {                    if (x == 0 || y == 0) {                        am[x][y] = 1;                    } else {                        am[x][y] = am[x-1][y-1] + 1;                    }                    if (am[x][y] > ret)                        ret = am[x][y];                }        }        return ret;    }};

时间复杂度：

emmm如何数组am的大小指定为[n][m]的话，复杂度应为O(nm) - - -n, m 分别为A, B的长度