LeetCode718. Maximum Length of Repeated Subarray

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718. Maximum Length of Repeated Subarray

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. 1 <= len(A), len(B) <= 1000
2. 0 <= A[i], B[i] < 100

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题目：数组A、B中相同子串的最大长度。

思路：动态规划。如果A，B中的某个子数组相同，假设是在位置A[i..]，B[j..]，用dp[i][j]表示以A[i]，B[j]开头的子数组相同部分的长度，那么当A[i]=B[j]时，以A[i],B[j]开头的子数组的相同部分的长度由A[i+1],B[j+1]开头的子数组的相同部分的长度决定（dp[i+1][j+1]）。状态转移矩阵为dp[i][j] = dp[i+1][j+1] + 1，每次更新dp[i][j]时，同时更新最大长度。

工程代码下载

class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        int lena = A.size();
        int lenb = B.size();
        vector<vector<int>> dp(lena+1, vector<int>(lenb+1));
        int res = 0;

        for(int i = lena-1; i >= 0 ; --i){
            for(int j = lenb-1; j >=0; --j){
                if(A[i] == B[j]){
                    dp[i][j] = dp[i+1][j+1] + 1;
                    res = max(res, dp[i][j]);
                }
            }
        }
        return res;
    }
};