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)

思路：这道题和2sum类型差不多，先排序，然后枚举第一个数，再设置头尾两个指针(j=i+1,k=n-1)，这样就相当于在查找两个数之和为目标值一样了。还有一点值得注意的是，去重判断。

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        int n=num.size();
        vector<vector<int> > result;
        result.clear();
        if(n<3)
            return result;
        sort(num.begin(),num.end());
        for(int i=0;i<n;i++)
        {
            if(i!=0 && num[i]==num[i-1])
                continue;
            int j=i+1,k=n-1;
            while(j<k)
            {
                if(j>i+1 && num[j]==num[j-1])
                {
                    j++;
                    continue;
                }
                if(k<n-1 && num[k]==num[k+1])
                {
                    k--;
                    continue;
                }
                int sum=num[i]+num[j]+num[k];
                if(sum==0)
                {
                    vector<int> temp;
                    temp.push_back(num[i]);
                    temp.push_back(num[j]);
                    temp.push_back(num[k]);
                    result.push_back(temp);
                    j++;
                }
                else if(sum<0)
                {
                    j++;
                }
                else
                {
                    k--;
                }
            }
        }
        return result;
    }
};