718. Maximum Length of Repeated Subarray【Medium】 动归

题目：找两个数组的最长公共子数组

方法：与寻找子串的算法一样：设A数组长度m，B数组长度n

用dp[1~m][1~n]存当前位置到前面某个位置可以得到的最长子串长度

dp[i][j] = 0                         --- A[i] != B[j]

               dp[i-1][j-1] + 1   --- A[i] == B[j]

最后返回dp里最大的值

较简单

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