Leetcode | Maximum Length of Repeated Subarray
来源:互联网 发布:废铁战士淘宝店怎么样 编辑:程序博客网 时间:2024/06/05 20:52
原题链接:https://leetcode.com/problems/maximum-length-of-repeated-subarray
原题意如下所示:
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
上述题意大概描述的就是给定两个数组A和B,然后找出它们之间公共的子数组,并且要求这个公共子数组的长度是所有公共子数组长度的最大值,最后返回这个最大的子数组长度。
由题意我们可以看出这个问题需要对这两个数组进行遍历,比较它们数组里面具体的数字,并用一些变量来记录它,一开始最容易想到的就是暴力破解的方法,但其实仔细思考后会发现解决这个问题的时候,用到的是动态规划的方法,即可以把它分解成一个个子问题,用dp[i][j]来记录A数组里面前i个数字和B数组里面前j个数字的当前记录的子字符串的长度,注意这里记录的只是当前的,并不是最大子字符串的长度,max才用来存最大子字符串的长度,当把两个数组都遍历完之后,max即是我们最后得到的答案。在这个过程中,还要处理一个子问题与原问题之间的关系,即怎样通过子问题的答案来获得原问题的解,这就要通过比较A[i-1]和B[j-1]是否相等,来确定dp[i][j]是等于1+dp[i-1][j-1]还是0,因为如果不能连续的话,就不能成为子数组了,所以不是的话直接就等于0,max在原max和dp[i][j]之间取一个最大值。
源代码如下所示:
class Solution {public: int findLength(vector<int>& A, vector<int>& B) { int a = A.size(); int b = B.size(); int dp[a+1][b+1]; int max = 0; for (int j = 0; j < b; j++) dp[0][j] = 0; for (int i = 0; i < a; i++) dp[i][0] = 0; for (int i = 1; i <= a; i++) { for (int j = 1; j <= b; j++) { if (A[i-1] == B[j-1]) { dp[i][j] = dp[i-1][j-1] + 1; if (max < dp[i][j]) max = dp[i][j]; } else { dp[i][j] = 0; } } } return max; }};
- leetcode 718( Maximum Length of Repeated Subarray)
- Leetcode 718. Maximum Length of Repeated Subarray
- Leetcode | Maximum Length of Repeated Subarray
- LeetCode #718 Maximum Length of Repeated Subarray
- Leetcode | Maximum Length of Repeated Subarray
- LeetCode:Maximum Length of Repeated Subarray
- [leetcode]Maximum Length of Repeated Subarray
- Leetcode 718. Maximum Length of Repeated Subarray
- leetcode 718. Maximum Length of Repeated Subarray
- [LeetCode]718. Maximum Length of Repeated Subarray
- [LeetCode]Maximum Length of Repeated Subarray
- [LeetCode] 718. Maximum Length of Repeated Subarray
- [leetcode] 718. Maximum Length of Repeated Subarray
- Leetcode 718 Maximum Length of Repeated Subarray
- LeetCode:Maximum Length of Repeated Subarray
- leetCode-Maximum Length of Repeated Subarray
- Leetcode:718. Maximum Length of Repeated Subarray
- Maximum Length of Repeated Subarray
- 10个实用的PHP正则表达式
- 欢迎使用CSDN-markdown编辑器
- jenkins 实现自动化构建(四)
- iPhone 8 销量如何?库克:哼
- 牛顿插值 C++ 和 Matlab实现
- Leetcode | Maximum Length of Repeated Subarray
- 《朗读者》:在历史的重负里前行
- java读取properties
- 打印图形:人人贷测试岗面试题(20171103)
- 求杨氏矩阵
- 模板:高精度
- 【服务器架构】网站架构要素和高性能架构
- HDU 2717- Catch That Cow
- iPhone SE不会淡出视野:苹果还需要它