problem:
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)
Solution: 先排序 然后左右夹逼 时间复杂度:O(n2)
class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> result; if(nums.size()<3) return result; //先排序 然后再左右夹逼 sort(nums.begin(),nums.end()); const int target=0; auto last=nums.end(); for(auto a=nums.begin();a<prev(last,2);a++) { auto b = next(a); auto c= prev(last); while(b<c) { if(*a+*b+*c<target) b++; else if(*a+*b+*c>target) --c; else{ result.push_back({*a,*b,*c}); ++b; --c; } } } return result; } };