leetcode | 3Sum
来源:互联网 发布:四大名捕 知乎 编辑:程序博客网 时间:2024/06/08 02:58
问题描述:点击打开链接 leetcode
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
- Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
- The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4}, A solution set is: (-1, 0, 1) (-1, -1, 2)
解析:
首先排序,然后左右夹逼,复杂度O(n^2) -- 从程序中可以看出
此方法可扩展到 k-sum (即求k个数的和等于target),先排序,然后做 k-2 次循环,在最内层循环中左右夹逼;时间复杂度为<math>max(O(nlogn),O(n^{k - 1}))</math>
(排序<math>O(nlogn)</math>, 循环<math>O(n^{k - 1})</math>
#include <iostream>#include <vector>#include <algorithm>using namespace std;class Solution {public:vector<vector<int>> threeSum(vector<int>& nums) {vector< vector<int> > result;if (nums.size() < 3)return result;sort(nums.begin(), nums.end());int target = 0;for (int i = 0; i < nums.size()-2; i++) {int j = i+1;int k = nums.size() - 1;if (i > 0 && nums[i] == nums[i-1])continue;while (j < k) {int sum = nums[i] + nums[j] + nums [k];if (sum < target) {j++;while (j < k && nums[j] == nums[j-1])j++;} else if (sum > target) {k--;while (j < k && nums[k] == nums[k+1])k--;} else {int t[] = {nums[i], nums[j], nums[k]};vector<int> temp(t, t+3);result.push_back(temp);cout << nums[i] << " "<< nums[j] << " " << nums[k] << endl;j++;k--;while (j < k && nums[j] == nums[j-1] && nums[k] == nums[k+1])//只有两个都等时,需要改变一个j++;}}}return result;}};int main(){int test1[] = {-1, 0, 1, 2, -1, -4};int test2[] = {-1, 0, 1};int test4[] = {-1, 0};int test3[] = {2, 3, -2, 0, -10, 4, 6};vector<int> nums(test1, test1+6);vector<int> nums2(test2, test2+3);vector<int> nums3(test3, test3+7);vector<int> nums4(test4, test4+2);Solution sol;sol.threeSum(nums);sol.threeSum(nums2);sol.threeSum(nums3);sol.threeSum(nums4);}
0 0
- 【Leetcode】3Sum (Sum)
- Leetcode:2Sum,3Sum
- 【Leetcode】3Sum Closest (Sum)
- leetcode 2 sum 3sum 4sum
- Leetcode 2SUM-3SUM-4SUM
- Leetcode-2sum,3sum,4sum
- leetcode 2 sum & 3 sum & 4 sum
- [LeetCode] 2Sum, 3Sum, 4Sum, 3SUm closet
- [LeetCode] K sum(2Sum、3Sum、4Sum)
- leetcode--sum集合:2sum,3sum,4sum
- leetcode --- 2 sum , 3 sum , 4 sum , k sum problem
- LeetCode: 3Sum
- LeetCode: 3 Sum Closest
- leetcode - 3 Sum
- leetcode - 3 sum closest
- leetcode 3Sum
- leetcode 3Sum Closest
- LeetCode: 3Sum
- Maven导出Project依赖的jar包
- Wireshark过滤规则
- 简单的android客户端servlet服务端的交互
- 正则表达式的简单梳理
- Android-drawable资源-InsetDrawable
- leetcode | 3Sum
- Java接口
- 《c++ primer》4 指针和const限定符
- nunit用法
- java中System.getProperty()方法详解
- 用户第一次拖动widget到桌面上,不用授权提示,直接安装
- QT学习(1)Windows环境下的串口编写
- Java Collection框架简略
- 收集android上开源的酷炫的交互动画和视觉效果:Interactive-animation