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LeetCode 454 4Sum II

Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that: \(0 \le i, j, k, l < n\) nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0 Solution 改写一下得到： \[nums1[i]+nums2[j]=-(nums3[k]+

LeetCode 454. 4Sum II

LeetCode 454. 4Sum II (四数相加 II) 题目 链接 https://leetcode.cn/problems/4sum-ii/ 问题描述 给你四个整数数组 nums1、nums2、nums3 和 nums4 ，数组长度都是 n ，请你计算有多少个元组 (i, j, k, l) 能满足： 0 <= i, j, k, l < n nums1[i] + nums2[j] + nums3[k] + nums4[l] ==

0454-leetcode算法实现之四数之和II-4sum-ii-python&golang实现

给你四个整数数组 nums1、nums2、nums3 和 nums4 ，数组长度都是 n ，请你计算有多少个元组 (i, j, k, l) 能满足： 0 <= i, j, k, l < n nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0 示例 1： 输入：nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2] 输出：2 解释： 两个元

Leetcode 454： 4Sum II

Leetcode 454： 4Sum II Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All in

18. 4Sum [Medium]

和3Sum一个思路，只是多一层循环来确定第2个数，之后还是用双指针找第3、4个数 2Sum一直到kSum都使用这样的套路解题 /** * 自己的代码 */ class Solution { public List<List<Integer>> fourSum(int[] nums, int target) { List<List<Integer>> ans = new ArrayList<

LeetCode C++ 454. 4Sum II【Hash Table/Sort/Two Pointers/Binary Search】

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the r

【LeetCode】454. 4Sum II 四数相加 II（Medium）（JAVA）

【LeetCode】454. 4Sum II 四数相加 II（Medium）（JAVA） 题目地址: https://leetcode.com/problems/4sum-ii/ 题目描述： Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. To make pr

454. 4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the rang

Leetcode 18. 4sum

题目描述 给出一个有n个元素的数组S，S中是否有元素a,b,c和d满足a+b+c+d=目标值？找出数组S中所有满足条件的四元组。 注意： 四元组（a、b、c、d）中的元素必须按非降序排列。（即a≤b≤c≤d） 解集中不能包含重复的四元组。 例如：给出的数组 S = {1 0 -1 0 -2 2}, 目标值 = 0.↵↵ 给

LC454. 4Sum II

   分析：暴力枚举复杂度为$O(n^4)$，不可行。将A，B能组成的和用map计数，查找 -C[i]-D[j] 在map中是否存在 time $O(n^2)$ space $O(n^2)$ class Solution { public: int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {

leetcode454 4Sum II

1 """ 2 Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. 3 To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. Al

【LeetCode】 18. 4Sum 四数之和（Medium）（JAVA）

【LeetCode】 18. 4Sum 四数之和（Medium）（JAVA） 题目地址: https://leetcode.com/problems/4sum/ 题目描述： Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadrupl

454. 4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range

454. 4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero. To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the rang

4Sum

题目描述： Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: Elements in a quadruplet (a,b,c,d) must be in

18. 4Sum[M]四数之和

题目 Given an array nums of n integers and an integer target, are there elements a, b, c and d in nums such that a+ b + c + d = target ? Find all unique quadruplets in the array which gives the sum of target. Note: The solution set must not contain duplicat

Leetcode 18. 4Sum

https://leetcode.com/problems/4sum/ #include<algorithm> #include<vector> using namespace std; class Solution { public: void _sum(vector<int>& nums,int l,vector<vector<int>> &ans,int target,vector<int> &x

leetcode 18-> 4Sum

  class Solution(object): def fourSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[List[int]] """ l = len(nums) nums.sort()

Leetcode 18. 4Sum

题目描述：找出数组中任意4个数和为target的组合，不得重复。 题目链接：Leetcode 18. 4Sum 思路：基于3Sum做待定，相当又套了一层循环。 当然，也可以暴力列举 3 3 个的情况来达到目的。 代码如下 // Author: Huahua class Solution { public: vector<vector<int>> fourSum(vector

18. 4Sum（js）

18. 4Sum Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: The solution set

[Swift]LeetCode454. 四数相加 II | 4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l)there are such that A[i] + B[j] + C[k] + D[l] is zero. To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range o

4Sum - LeetCode

目录 题目链接 注意点 解法 小结 题目链接 Merge Two Sorted Lists - LeetCode 注意点 和3Sum那道题一样 解法 解法一：在3Sum的基础上再加一层循环即可。时间复杂度为O(n^3) class Solution { public: using num_t = vector<int>; vector<vector<int>> fourSum(vector<i