LeetCode Weekly Contest 56 第三题

题目

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1. Maximum Length of Repeated Subarray My SubmissionsBack to Contest
   User Accepted: 778
   User Tried: 1121
   Total Accepted: 744
   Total Submissions: 1920
   Difficulty: Medium
   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

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分析

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题目意思就是找最大的重叠项

暴力模拟吧，这题可以动态规划都是O（M*N）题解里的暴力和我的暴力不太一样

我们用类似滑动校验的思想

先从头匹配

接着固定一端右移一项匹配

知道一个的末端

这时的解是不完全的

我们需要再换一个固定右移

详细的看代码

动态规划的开个二维的数组可以ac

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代码

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class Solution {public:    int findLength(vector<int>& A, vector<int>& B) {        int ans =0 ;         for( int i=0 ; i<A.size() ;i++){            int length =0 ;             for( int j =0 , k = i ; j<B.size() && k < A.size() ; j++,k++){                if( B[j] == A[k ]){                    length++;                    ans = max( ans ,length );                }                else{                    length = 0 ;                }            }        }        for( int i=0 ; i<B.size() ;i++){            int length =0 ;             for( int j =0 , k = i ; j<A.size() && k < B.size() ; j++,k++){                if( A[j] == B[k ]){                    length++;                    ans = max( ans ,length );                }                else{                    length = 0 ;                }            }        }        return ans ;    }};

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时间复杂度

O(M∗N)

空间复杂度

O(1)