LWC 56:718. Maximum Length of Repeated Subarray
来源:互联网 发布:淘宝便利店 编辑:程序博客网 时间:2024/06/04 19:39
LWC 56:718. Maximum Length of Repeated Subarray
传送门:718. Maximum Length of Repeated Subarray
Problem:
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:
和上周一样,还是一道动态规划题,如果对动规不熟悉,可以采用递归+记忆化搜索,本题亦如此。首先判断末位两个元素是否相等,不相等把A的问题规模变成n-1,和B求出最长subarray长度,或者把B的问题规模变成m-1,和A求出最长subarray。
遇到相等该如何操作,自然在子问题n-1,和m-1基础上+1,(问题的定义如下:给定A和B的末区间i和j,返回A[0,…,i]和B[0,…,j]中以i和j结尾的最长subArray长度),
所以有如下递归式:
f(A, B, i, j) = 1 + f(A, B, i - 1, j - 1); if (A[i] == B[j])or:f(A, B, i, j) = 0;
此时就可以用数组dp记录每个状态的最长subArray长度,并在过程中不断更新max。
代码如下:
public int findLength(int[] A, int[] B) { max = 0; int n = A.length; int m = B.length; dp = new int[n + 1][m + 1]; for (int i = 0; i < n + 1; ++i) { for (int j = 0; j < m + 1; ++j) { dp[i][j] = -1; } } dfs(A, B, n - 1, m - 1); return max; } int[][] dp; int max = 0; int dfs(int[] A, int[] B, int n, int m) { if (n == -1 || m == -1) return 0; if (dp[n][m] >= 0) return dp[n][m]; dfs(A, B, n - 1, m); dfs(A, B, n, m - 1); if (A[n] == B[m]) { int ans = 1; int sub = dfs(A, B, n - 1, m - 1); max = Math.max(max, ans + sub); dp[n][m] = ans + sub; return ans + sub; } else { dp[n][m] = 0; return 0; } }
注意此处,0也是可能的一种答案,所以DP初始化为-1,否则超时。
递推版本(动态规划):
public int findLength(int[] A, int[] B) { int n = A.length; int m = B.length; int max = 0; int[][] dp = new int[n + 1][m + 1]; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= m; ++j) { if (A[i - 1] == B[j - 1]) { dp[i][j] = dp[i - 1][j - 1] + 1; } else { dp[i][j] = 0; } max = Math.max(dp[i][j], max); } } return max; }
阅读全文
0 0
- LWC 56:718. Maximum Length of Repeated Subarray
- Leetcode 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- 718.Maximum Length of Repeated Subarray(M)
- 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- Leetcode 718. Maximum Length of Repeated Subarray
- leetcode 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- [LeetCode]718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- 718. Maximum Length of Repeated Subarray
- SpringMVC学习之简单示例(非注解)
- Python之正则表达式入门
- sleep和wait的区别
- 标称型和数值型数据
- oracle查看被锁的表和解锁
- LWC 56:718. Maximum Length of Repeated Subarray
- 修改IP地址失败的解决方法
- Java初学的一些笔记
- 数据结构学习记录-树的基本术语
- Redis 发布订阅
- Spring Struts2 Hibernate
- 46. Permutations
- Android逆向之旅---Android手机端破解神器MT的内购VIP功能破解教程
- 二分+SPFA