学步园

Problem:

Given an array S of n integers,
are there elements a, b, c,
and d in S such
that a + b + c + d =
target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d)
  must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

个人解法：

1. 对数组排序
2. 确定四元数中的两个
3. 遍历剩余数组确定两外两个

算法时间复杂度为O(n^3)。

网上看到一个时间复杂度O(n^2)，空间复杂度为O(n)的解法：采用分治思想，先对数组预处理，求的元素两两之和，然后采用2Sum算法思想遍历数组求的四个数字之和。

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int nSize = num.size();
        vector< vector<int> > result;
        if (nSize < 4) return result;
        
        sort(num.begin(), num.end());
        vector<int> mid(4);
        set<string> isExit;
        for (int i = 0; i < nSize - 3; ++i)
        {
            mid[0] = num[i];
            for (int j = i + 1; j < nSize - 2; ++j)
            {
                mid[1] = num[j];
                int l = j + 1;
                int r = nSize - 1;
                int sum = target - num[i] - num[j];
                while(l < r)
                {
                    int tmp = num[l] + num[r];
                    if (sum == tmp)
                    {
                        string str;
                        str += num[i];
                        str += num[j];
                        str += num[l];
                        str += num[r];
                        set<string>::iterator itr = isExit.find(str);
                        if (itr == isExit.end())
                        {
                            isExit.insert(str);
                            mid[2] = num[l];
                            mid[3] = num[r];
                            result.push_back(mid);
                        }
                        ++l;
                        --r;
                    }
                    else if(sum > tmp)
                        ++l;
                    else
                        --r;
                }
            }
        }

        return result;
    }
};