718. Maximum Length of Repeated Subarray

最新推荐文章于 2021-05-10 06:16:38 发布

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本文介绍了一种通过双层循环对比的方法来寻找两个整数数组中出现的最长相同子数组的有效解决方案。通过逐元素对比，该算法能找出最长的重复子数组。

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

思路：使用循环去比较

        1 2 3 2 1       1 2 3 2 1      1 2 3 2 1
3 2 1 4 7         3 2 1 4 7     ...        3 2 1 4 7

class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        int lenA = A.size();
        int lenB = B.size();
        int len = 0;
        for(int i = lenB - 1; i >= 0; i--){          
            int tmp = 0;
            for(int j = i, m = 0; j < lenB && m < lenA; j++, m++){
                if(B[j] == A[m]){
                    tmp++;
                } else {
                    len = max(len, tmp);
                    tmp = 0;
                }
            }

            len = max(len, tmp);
        }
        for(int i = 1; i < lenA; i++){
            int tmp = 0;
            for(int j = 0, m = i; j < lenB && m < lenA; j++, m++){
                if(B[j] == A[m]){
                    tmp++;
                } else {
                    len = max(len, tmp);
                    tmp = 0;
                }                    
            }
            len = max(len, tmp);
        }
        return len;
    }
};