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)
这题做得真是郁闷。。。。
C++实现代码：
#include<iostream> #include<algorithm> #include<vector> using namespace std; class Solution { public: vector<vector<int> > threeSum(vector<int> &num) { if(num.empty()) return vector<vector<int> >(); sort(num.begin(),num.end()); vector<vector<int> > ret; int n=num.size(); int i; for(i=0; i<n-2; i++) { //只保留第一个不重复的，其余的都删了 if(i>=1&&num[i]==num[i-1]) continue; int target=-num[i]; int left=i+1; int right=n-1; vector<int> tmp; while(left<right) { //删除重复元素 if(right>left&&right<n-1&&num[right+1]==num[right]) right--; else if(num[left]+num[right]==target) { tmp={num[i],num[left],num[right]}; ret.push_back(tmp); left++; right--; } else if(num[left]+num[right]<target) left++; else if(num[left]+num[right]>target) right--; } } return ret; } }; int main() { vector<int> vec= {-2,0,1,1,1,1,1,2}; Solution s; vector<vector<int> > result=s.threeSum(vec); for(auto a:result) { for(auto v:a) cout<<v<<" "; cout<<endl; } cout<<endl; }
