718. Maximum Length of Repeated Subarray

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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

题目大意

给定两个vector数组，找出其中存在的最长的共同子串。

解题思路

用动态规划的思路，以空间换取时间。创建一个二维数组lenRecord，来记录共同子串的长度。当A[i] 和 B[j] 相等时，以A[i - 1] 和 B[j - 1] 结束的子串可以增长1，即lenRecord[i - 1][j - 1]+1；如果不相等，则该位置lenRecord[i][j]置为零 。最后，将得到的lenRecord[i][j]与max相比将，取较大值。以此类推，可以获得最大共同子串的长度。

算法复杂度

O(N2)

代码实现

class Solution {public:    int findLength(vector<int>& A, vector<int>& B) {        int lenA = A.size(), lenB = B.size();        int maxLen = 0;        vector<int> col(lenB+1, 0);        vector<vector<int> > lenRecord(lenA+1, col);        for (int i = 1; i < lenA+1; ++i) {            for (int j = 1; j < lenB+1; ++j) {                lenRecord[i][j] = (A[i-1] == B[j-1]) ? (lenRecord[i-1][j-1] + 1) : 0;                maxLen = max(maxLen, lenRecord[i][j]);            }        }        return maxLen;    }};