学步园

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：先确定一个数字，后两个数字用夹逼方式来扫描实现。通过判断并去重

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > result;
        if(num.size() < 3)
            return result;
            
        sort(num.begin(), num.end());
        const int target = 0;
        
        int a,b,c;
        int last_index = num.size() - 2;
        for(a = 0; a < last_index; ++a)
        {
            b = a + 1;
            c = num.size() - 1;
            if((a != 0)&&(num[a-1] == num[a]))//去重
                continue;
            while(b < c)
            {
                if((num[a] + num[b] + num[c]) < target)
                    ++b;
                else if((num[a] + num[b] + num[c]) > target)
                    --c;
                else{
                    if((b != (a+1))&&(num[b-1] == num[b])){
                        ++b;
                        continue;
                    }    
                    if((c != (num.size()-1))&&(num[c] == num[c+1])){ //去重
                        --c;
                        continue;
                    }    
                    result.push_back({num[a], num[b], num[c]});
                    ++b;
                    --c;
                }
            }
        }
        
        return result;
    }
};