Leetcode: 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)
递归:
void threeSumHelper(vector<int> &num, int index, vector<int> &path, int cursum, vector<vector<int>> &res) { if(path.size() >=3) { if(path.size() == 3 && cursum == 0) { res.push_back(path); return; }else return; } for(int i = index; i < num.size(); i++) { path.push_back(num[i]); cursum += num[i]; threeSumHelper(num,i+1,path,cursum,res); path.pop_back(); cursum -= num[i]; while(i < num.size()-1 && num[i] == num[i+1])i++; } } vector<vector<int> > threeSum(vector<int> &num) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<vector<int>> res; if(num.size()<3)return res; sort(num.begin(), num.end()); vector<int> path; threeSumHelper(num,0,path,0,res); return res; }
2sum变形:
vector<vector<int> > threeSum(vector<int> &num) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<vector<int>> res; if(num.size()<3)return res; set<vector<int>> tmpres; sort(num.begin(), num.end()); for(int i = 0; i < num.size(); i++) { int target = - num[i]; int begin = i+ 1; int end = num.size()-1; while(begin < end) { int sum = num[begin] + num[end]; if( sum == target) { vector<int> tmp; tmp.push_back(num[i]); tmp.push_back(num[begin]); tmp.push_back(num[end]); tmpres.insert(tmp); begin++; end--; }else if(sum < target) begin++; else end--; } } set<vector<int>>::iterator it = tmpres.begin(); for(; it != tmpres.end(); it++) res.push_back(*it); return res; }
补充:软件开发 , C++ ,