[leetcode] 4Sum

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)

即给定一个数组S，输出S中的四个元素使得之和为0，且每个元组中元素从小到大的排列，且不重不漏。

解法和3Sum类似，使用四个指针i，j，p，q来遍历数组。其中pq分别从首和尾开始。注意判重方法与3Sum相同。

代码如下：

class Solution {
    public:
         vector<vector<int> > fourSum(vector<int> &num, int target) {
            vector<vector<int> > res;
            if( num.size() < 4 )
            {
                return res;
            }
            vector<int> arr = num;
            sort(arr.begin(), arr.end());
            for( int i = 0 ; i < arr.size()-3 ; i++ )
            {
                if(i > 0 && arr[i] == arr[i-1])
                  continue;
                for( int j = i+1 ; j < arr.size()-2 ; j++ )
                {
                    if(j > i+1 && arr[j] == arr[j-1])
                      continue;
                    int p = j+1;
                    int q = arr.size()-1;
                    while(p < q)
                    {
                        if (p > j + 1 && arr[p] == arr[p-1])
                        {
                            p++;
                            continue;
                        }
                        if (q < arr.size() - 1 && arr[q] == arr[q+1])
                        {
                            q--;
                            continue;
                        }
                        int sumtmp  = arr[i] + arr[j] + arr[p] + arr[q] ;
                        if (sumtmp == target)
                        {
                            vector<int> tmp;
                            tmp.push_back(arr[i]);
                            tmp.push_back(arr[j]);
                            tmp.push_back(arr[p]);
                            tmp.push_back(arr[q]);
                            res.push_back(tmp);
                            p++;
                        }
                        else if( sumtmp < target )
                        {
                            p++;
                        }
                        else
                        {
                            q--;
                        }
                    }
                }

            }
            return res;
        }
};