718. Maximum Length of Repeated Subarray

标签（空格分隔）： leetcode

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1.原题

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

2. 分析

• 采用动态规划的方法， 有2个数组， 以dp[i][j]表示以A[ i ] 和 B[ j ]结尾的最长的字符串
• 如果A[i] == B[j] 那么以A[ i ] 和 B[ j ]结尾的最长的字符串等于A[ i - 1][ j - 1]的最长的字符串长度加上这一位， dp[i][j]=dp[i−1][j−1]+1
• 考虑越界问题以及初始化无意义的边缘
• 考虑怎么初始化二维的动态数组才不会出错。

3. 代码

class Solution {public:    int findLength(vector<int>& A, vector<int>& B) {        int len1 = A.size();        int len2 = B.size();        if (len1 == 0 || len2 == 0)            return 0;        int **dp = new int*[len1 + 1];        int max = -1;        for (int i = 0; i <= len1; i++) {            dp[i] = new int[len2 + 1];            for (int j = 0; j <= len2; j++) {                if (i == 0 || j == 0) {                    dp[i][j] = 0;                }                else {                    if (A[i - 1] == B[j - 1]) {                        dp[i][j] = dp[i - 1][j - 1] + 1;                    }                    else {                        dp[i][j] = 0;                    }                    if (max < dp[i][j]) {                        max = dp[i][j];                    }                }            }        }        return max;    }};