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3Sum
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)class Solution {public: vector<vector<int> > threeSum(vector<int> &num) { vector<vector<int> > ans; if(num.size() < 3) return ans; sort(num.begin(), num.end()); int pos = 0; while(num[pos]<0){pos++;} int zeroind = 0; while(num[pos+zeroind] == 0) zeroind++; if(zeroind >= 3) { vector<int> tmp; tmp.push_back(0); tmp.push_back(0); tmp.push_back(0); ans.push_back(tmp); } int i=0; int j=pos; while(i<pos){ j=pos; while(j<num.size()){ int c = 0 - (num[i] + num[j]); if(c < 0 && c >= num[i]){ for(int k=i+1; k<pos; k++){ if(num[k] == c){ vector<int> tmp; tmp.push_back(num[i]); tmp.push_back(c); tmp.push_back(num[j]); ans.push_back(tmp); break; } } } if(c>=0 && c>=num[j]){ for(int k=j+1; k<num.size(); k++){ if(num[k] == c){ vector<int> tmp; tmp.push_back(num[i]); tmp.push_back(num[j]); tmp.push_back(c); ans.push_back(tmp); break; } } } while(num[j] == num[j+1]) j++; j++; } while(num[i] == num[i+1]) i++; i++; } return ans; }};
3Sum
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