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：
接着2Sum的思路，只需枚举第一个数，然后判断后两个数和是否是第一个的负数即可。
在判断重复的时候卡了挺久，其实只需保证每次新的判断是和上次不同的子问题就行。

代码：

    vector<vector<int> > threeSum(vector<int> &num) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<vector<int> > result;
        vector<int> tmp;
        result.clear();
        tmp.clear();
        int len = num.size();
        int i, j, k;
        sort(num.begin(), num.end());
        for(i = 0; i < len; i++){
            while(i > 0 && num[i] == num[i-1])
                i++;
            j = i+1;
            k = len-1;
            while(j < k){
                while(j < k && num[j] == num[j-1] && j > i+1)
                    j++;
                while(j < k && num[k] == num[k+1] && k < len-1)
                    k--;
                if(j >= k)
                    break;
                int t = num[j] + num[k];
                if(t < 0 - num[i])
                    j++;
                else if(t > 0 - num[i])
                    k--;
                else{
                    tmp.push_back(num[i]);
                    tmp.push_back(num[j]);
                    tmp.push_back(num[k]);
                    result.push_back(tmp);
                    j++;
                    k--;
                    tmp.clear();
                }
            }
        }
        return result;
    }