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)

思路：

思路一是递归的方式用01背包问题来解决。但是需要检查冲突，时间开销比较大。思路二是三层循环，首先对数组排序，前两层是完全的循环，最后一层是二分查找，考虑到有连续的相同值，时间上还可以再优化。

代码：

思路一：

class Solution {
public:
    vector<vector<int> >* v;
    vector<vector<int> > threeSum(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        v = new vector<vector<int>>();
        if(num.size() < 3)
        {
            return *v;
        }
        
        sort(num.begin(), num.end());
        
        vector<int> res;
        add(num, 0, res);
        
        return *v;
    }
    
    void add(vector<int> &num, int i, vector<int> res)
    {
        if(i < num.size())
        {
            if(res.size() == 2 && res[0] + res[1] + num[i] == 0)
            {
                vector<int> tmp;
                tmp.push_back(res[0]);
                tmp.push_back(res[1]);
                tmp.push_back(num[i]);
                
                bool duplicate = false;
                for(int k = 0; k < v->size(); k++)
                {
                    if(v->at(k) == tmp)
                    {
                        duplicate = true;
                        break;
                    }
                }
                if(!duplicate)
                    v->push_back(tmp);
            }
            else if(res.size() == 2 && res[0] + res[1] + num[i] != 0)
            {
                add(num, i+1, res);
            }
            else if(res.size()<2)
            {
                add(num, i+1, res);
                res.push_back(num[i]);
                add(num, i+1, res);
            }
        }
    }
};

思路二：

class Solution {
public:
    vector<vector<int> >* v;
    vector<vector<int> > threeSum(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        v = new vector<vector<int>>();
        if(num.size() < 3)
        {
            return *v;
        }
        
        sort(num.begin(), num.end());
        
        for(int i = 0; i < num.size()-2; i++)
        {
            if(i > 0 && num[i] == num[i-1])
            {
                continue;
            }
            for(int j = i+1; j < num.size()-1; j++)
            {
                if(j > i+1 && num[j] == num[j-1])
                {
                    continue;
                }
                int left = j+1;
                int right = num.size()-1;
                while(left<=right)
                {
                    int mid = left + ((right-left)>>1);
                    if(num[i] + num[j] + num[mid] == 0)
                    {
                        vector<int> tmp;
                        tmp.push_back(num[i]);
                        tmp.push_back(num[j]);
                        tmp.push_back(num[mid]);
                        
                        v->push_back(tmp);
                        break;
                    }
                    else if(num[i] + num[j] + num[mid] > 0)
                    {
                        right = mid - 1;
                    }
                    else
                    {
                        left = mid + 1;
                    }
                }
            }
        }
        
        return *v;
    }
};