leetcode习题解答： 15. 3Sum

难度：medium

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: 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]]

思路：可以用3重循环来暴力破解，但是这样的话复杂度是O(n^3)

我们可以先排序数据，然后定住一个数，剩下的数做2Sum中的两端逼近的算法，这样算法的复杂度就变成了O(n^2)

代码：

#include <algorithm>#include <set>class Solution {public:    vector<vector<int>> threeSum(vector<int>& num) {        vector<int> nums = num;        vector<vector<int>> result;        set<vector<int>> s;        sort(nums.begin(),nums.end());        for (int i = 0; i < nums.size(); i++){            int value = nums[i];            int begin = i+1;            int end = nums.size()-1;            while (begin < end){                if (nums[begin]+nums[end] == -value){                    vector<int> temp({value,nums[begin],nums[end]});                    if (s.find(temp) == s.end()){                        result.push_back(temp);                        s.insert(temp);                    }                    end--;                    begin++;                } else if (nums[begin]+nums[end] > -value){                    end--;                } else {                    begin++;                }            }        }        return result;    }};

set是用来去重的