[LeetCode]718. Maximum Length of Repeated Subarray

• descrption
  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: 3Explanation: The repeated subarray with maximum length is [3, 2, 1].Note:1 <= len(A), len(B) <= 10000 <= A[i], B[i] < 100

• 解题思路
  求最长子串长度，利用动态规划算法。找到动态规划的状态和状态转移方程就可以解决问题。令C[i][j]表示以A[i]和b[j]结尾的最长字串长度，如果A[i]不等于B[j]，那么C[i][j]就等于0，如果相等，那么C[i][j]就等于以A[i-1]和B[j-1]结尾的最长子串长度+1。

C[i][j] = A[i]==B[j]?C[i−1][j−1]+1:0

• 代码如下

class Solution {public:    int findLength(vector<int>& A, vector<int>& B) {        vector<vector<int>> C(A.size()+1, vector<int>(B.size()+1, 0));        int ans = 0;        for(int i = 1; i < A.size() + 1; i++){            for(int j = 1; j < B.size() + 1; j ++){                if(A[i-1] == B[j-1]) {                    C[i][j] = C[i-1][j-1] + 1;                }                if(ans < C[i][j]){                    ans = C[i][j];                }             }        }        return ans;    }};

原题地址

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