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
最长重复子数组。
题目很好懂。这题可以同1143题一起做,几乎一模一样。
既然是找子数组,而且暴力解一看复杂度就比较高,既要扫描两个数组又要找相同的部分,所以思路是尝试用动态规划来解决问题。动态规划的二维数组 dp[i][j] 的含义是以 A[i] 和 B[j] 为结尾的公共子数组的最长长度。
- 当A[i] != B[j]时,dp[i][j] = 0, 因为以A[i]和B[j]结尾的公共子串不存在,因为他们最后一个字母不同
- 当A[i] == B[j]时,dp[i][j] = dp[i-1][j-1] + 1, 因为A[i]和B[j]相等,以他们为结尾的最长公共子串的长度就是以A[i-1]和B[j-1]结尾的最长公共子串长度 + 1
时间O(mn)
空间O(1)
Java实现
1 class Solution { 2 public int findLength(int[] A, int[] B) { 3 int m = A.length; 4 int n = B.length; 5 int[][] dp = new int[m + 1][n + 1]; 6 int res = 0; 7 for (int i = 1; i <= A.length; i++) { 8 for (int j = 1; j <= B.length; j++) { 9 dp[i][j] = A[i - 1] == B[j - 1] ? dp[i - 1][j - 1] + 1 : 0; 10 res = Math.max(res, dp[i][j]); 11 } 12 } 13 return res; 14 } 15 }
JavaScript实现
1 /** 2 * @param {number[]} A 3 * @param {number[]} B 4 * @return {number} 5 */ 6 var findLength = function (A, B) { 7 let m = A.length; 8 let n = B.length; 9 let dp = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0)); 10 let res = 0; 11 for (let i = 1; i <= m; i++) { 12 for (let j = 1; j <= n; j++) { 13 dp[i][j] = A[i - 1] == B[j - 1] ? dp[i - 1][j - 1] + 1 : 0; 14 res = Math.max(res, dp[i][j]); 15 } 16 } 17 return res; 18 };
相关题目
718. Maximum Length of Repeated Subarray