718. Maximum Length of Repeated Subarray

问题:

给定两个数组,求两个数组中最长公共子数组的长度。

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

  

解法:

DP动态规划,

dp[i][j]代表到A到第i个数B到第j个数为止,连续公共子数组到长度。(且到此位,两个公共数组还在相等比较下去中)

动态转移方程:

if(A[i]==B[j]) {
    dp[i+1][j+1]=dp[i][j]+1;
} else {
    dp[i+1][j+1]=0;
} 

 

边界值:dp[0][j] or dp[i][0] =0

例如,有以下两个数列,则构建到dp为以下memo数组到样子:

718. Maximum Length of Repeated Subarray

 

 参考代码:

 1 class Solution {
 2 public:
 3     //dp[i][j]=A[i]==B[j]?dp[i-1][j-1]+1:0;
 4     int findLength(vector<int>& A, vector<int>& B) {
 5         int n = A.size();
 6         int m = B.size();
 7         int res=0;
 8         int dp[n+1][m+1];
 9         memset(dp,0,sizeof(dp));
10         for(int i=1; i<=n; i++){
11             for(int j=1; j<=m; j++){
12                 if(A[i-1]==B[j-1]) dp[i][j]=dp[i-1][j-1]+1;
13                 else dp[i][j]=0;
14                 res=max(res, dp[i][j]);
15             }
16         }
17         return res;
18     }
19 };

 

  

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