LeetCode题解：3Sum

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：

先回忆一下2sum问题的解法。对从小到大排序的数组，在数组前后各放一个指针。如果指针对应数的和大于给定值，则右移最左边的指针，反之左移最右边的指针，直到和是给定值或者两个指针碰撞。3Sum可以首先固定一个数，然后进行2Sum。

如果需要unique的解法，一种方法是先给出所有解，然后做一个std::unique，另一种方法是记录之前指针的值，然后进行新的循环时检测指针对应的值是否更新。

题解：

class Solution
{
public:
    vector<vector<int>> ret;
    int head;
    
    void two_sum (const vector<int>& v,
                  vector<int>::const_iterator c1,
                  int expectation)
    {
        if (v.cend() - c1 <= 1)
            return;
            
        vector<int>::const_iterator c2 = v.cend() - 1;
        while (c1 < c2)
        {
            int twos = *c1 + *c2;
            if (twos == expectation)
            {
                ret.emplace_back (vector<int> {{head, *c1, *c2}});
                
                int cc1 = *c1;
                while (c1 < c2 && cc1 == *c1) ++c1;
            }
            else if (twos < expectation)
                ++c1;
            else 
                --c2;
        }
    }
    
    vector<vector<int> > threeSum (vector<int>& num)
    {
        ret.clear();
        
        if (num.empty())
            return ret;
            
        sort (begin (num), end (num));
        head = num.front();
        for (auto n = num.cbegin(); n != num.cend(); ++n)
        {
            if (n != num.cbegin() && *n == head)
                continue;
            head = *n;
            two_sum (num, n + 1, -*n);
        }
        return ret;
    }
};