3Sum

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：

先回憶一下2sum問題的解法。對從小到大排序的陣列，在陣列前後各放一個指標。如果指標對應數的和大於給定值，則右移最左邊的指標，反之左移最右邊的指標，直到和是給定值或者兩個指標碰撞。3Sum可以首先固定一個數，然後進行2Sum。

如果需要unique的解法，一種方法是先給出所有解，然後做一個std::unique，另一種方法是記錄之前指標的值，然後進行新的迴圈時檢測指標對應的值是否更新。

題解：

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;
    }
};