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718. 最长重复子数组

English Version

题目描述

给两个整数数组 nums1 和 nums2 ，返回 两个数组中 公共的 、长度最长的子数组的长度 。

 

示例 1：

输入：nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
输出：3
解释：长度最长的公共子数组是 [3,2,1] 。

示例 2：

输入：nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
输出：5

 

提示：

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 100

解法

方法一：动态规划

Python3

class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        ans = 0
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if nums1[i - 1] == nums2[j - 1]:
                    dp[i][j] = 1 + dp[i - 1][j - 1]
                    ans = max(ans, dp[i][j])
        return ans

Java

class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;
        int[][] dp = new int[m + 1][n + 1];
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (nums1[i - 1] == nums2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    ans = Math.max(ans, dp[i][j]);
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int findLength(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (nums1[i - 1] == nums2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    ans = max(ans, dp[i][j]);
                }
            }
        }
        return ans;
    }
};

Go

func findLength(nums1 []int, nums2 []int) int {
	m, n := len(nums1), len(nums2)
	dp := make([][]int, m+1)
	for i := range dp {
		dp[i] = make([]int, n+1)
	}
	ans := 0
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if nums1[i-1] == nums2[j-1] {
				dp[i][j] = dp[i-1][j-1] + 1
				if ans < dp[i][j] {
					ans = dp[i][j]
				}
			}
		}
	}
	return ans
}

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