Maximum Length of Repeated Subarray
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Maximum Length of Repeated Subarray
题目描述
解题思路
题目属于动态规划一栏,自然会下意识通过动态规划方面的思想去想对策,动态规划算法的核心便是状态与状态转移方程,即当前的状态需要依靠前一个状态的值。而在本题中我完全可以用C[i][j] 来表示各自数组分别以A[i] 和 B[j] 结尾的最长公共子串的长度,如果A[i]==B[j],C[i][j]=C[i−1][j−1]+1。
C++代码实现
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 max = 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; } max = max > C[i][j] ? max : C[i][j]; } } return max; }};
题目样例测试结果
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