Week11

Problem–Medium–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

题目解析：

 这是一道典型的动态规划题目，即两段字符串中找最长公共子串。首先，我们需要寻找原问题的子问题，原问题求两段长度分别为ｍ，ｎ的字符串的最长公共子序列，那么，我们可以先求前ｎ-1与m-1长的字符串的最长公共子序列，看以第n-1与m-1结尾的序列的最长公共子序列长度是多少，假设第ｎ与m个字符也相等，那么只需在前面的基础上加１即可，那么我们就得到了一个状态转移方程： 初始化：dp[i][j] = 0
(i=0||j=0) 转移方程：dp[i][j] = dp[i – 1][j – 1] + 1
(A[i] = B[j])
  dp[i][j] = dp[i – 1][j – 1]
(A[i] != B[j])

代码：

class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        int ** dp = new int *[A.size() + 1];
        for (int i = 0; i <= A.size(); i++) {
            dp[i] = new int[B.size() + 1];
        }
        for (int i = 0; i <= A.size(); i++) {
            for (int j = 0; j <= B.size(); j++) {
                dp[i][j] = 0;
            }
        }
        int Max = 0;
        for (int i = 1; i <= A.size(); i++) {
            for (int j = 1; j <= B.size(); j++) {
                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];
                }
                //cout << "dp " << i << " " << j << " = " <<dp[i][j] << endl;
            }
        }
        return Max;
    }
};