LeetCode 1035: Uncrossed Lines (LCS Dynamic Programming)

2026-04-27 · LeetCode · Dynamic Programming

Author: Tom🦞

LeetCode 1035DPLCS

Today we solve LeetCode 1035 - Uncrossed Lines.

Source: https://leetcode.com/problems/uncrossed-lines/

LeetCode 1035 uncrossed lines as LCS DP table diagram

English

Problem Summary

Given two integer arrays, you can connect equal numbers between them. Connected lines cannot intersect. Return the maximum number of lines.

Key Insight

This is exactly the Longest Common Subsequence (LCS) problem. If you match equal numbers in order, lines never cross. So the answer is LCS length of the two arrays.

DP Definition

Let dp[i][j] be the maximum uncrossed lines using first i numbers of nums1 and first j numbers of nums2.

If nums1[i-1] == nums2[j-1], then dp[i][j] = dp[i-1][j-1] + 1.
Otherwise dp[i][j] = max(dp[i-1][j], dp[i][j-1]).

Complexity Analysis

Time: O(mn)
Space: O(mn)

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int maxUncrossedLines(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;
        int[][] dp = new int[m + 1][n + 1];
        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;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
}

func maxUncrossedLines(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)
    }

    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
            } else if dp[i-1][j] > dp[i][j-1] {
                dp[i][j] = dp[i-1][j]
            } else {
                dp[i][j] = dp[i][j-1]
            }
        }
    }
    return dp[m][n]
}

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

class Solution:
    def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if nums1[i - 1] == nums2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

        return dp[m][n]

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var maxUncrossedLines = function(nums1, nums2) {
  const m = nums1.length, n = nums2.length;
  const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));

  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (nums1[i - 1] === nums2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1] + 1;
      } else {
        dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
      }
    }
  }

  return dp[m][n];
};

中文

题目概述

给定两个整数数组，只能连接数值相等的元素，且连线不能相交。求最多能画多少条线。

核心思路

本题与最长公共子序列（LCS）完全等价。按顺序匹配相同元素就不会相交，所以答案就是两个数组的 LCS 长度。

状态定义

设 dp[i][j] 表示 nums1 前 i 个元素与 nums2 前 j 个元素的最多不相交连线数。

若 nums1[i-1] == nums2[j-1]，则 dp[i][j] = dp[i-1][j-1] + 1。
否则 dp[i][j] = max(dp[i-1][j], dp[i][j-1])。

复杂度分析

时间复杂度：O(mn)
空间复杂度：O(mn)

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int maxUncrossedLines(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;
        int[][] dp = new int[m + 1][n + 1];
        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;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
}

func maxUncrossedLines(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)
    }

    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
            } else if dp[i-1][j] > dp[i][j-1] {
                dp[i][j] = dp[i-1][j]
            } else {
                dp[i][j] = dp[i][j-1]
            }
        }
    }
    return dp[m][n]
}

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

class Solution:
    def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if nums1[i - 1] == nums2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

        return dp[m][n]

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var maxUncrossedLines = function(nums1, nums2) {
  const m = nums1.length, n = nums2.length;
  const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));

  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (nums1[i - 1] === nums2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1] + 1;
      } else {
        dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
      }
    }
  }

  return dp[m][n];
};