LeetCode 3Sum

题目：

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)

思想：首先将数组进行排序，复杂度O(nlgn)；然后固定一个数，问题就退化成2sum，复杂度O(n^2)。

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
    	int len = num.size();
        vector<vector<int>> res;
        //O(nlgn)
        sort(num.begin(), num.end());
        vector<int> tmp(3, INT_MIN);
        //O(n^2)
        for(int i = 0; i < len-2; i++) {
            //之前计算过的无需再计算
        	if(num[i] == tmp[0])
        		continue;
        	tmp[0] = num[i];
        	int begin = i+1, end = len-1, target = 0 - num[i];
			while(begin < end) {
				int sum = num[begin] + num[end];
				if(sum == target) {
					tmp[1] = num[begin];
					tmp[2] = num[end];
					res.push_back(tmp);
					begin++;
					end--;
					continue;	
				}
				else if(target > sum)
					begin++;
				else
					end--;
			} 
        }
        //使用set的目的是去重
        set<vector<int>> ans(res.begin(), res.end());
        res.clear();
        res.insert(res.begin(), ans.begin(), ans.end());
        return res; 
    }
};