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：
1 class Solution { 2 public: 3 vector<vector<int> > fourSum(vector<int> &num, int target) { 4 vector<vector<int> > res; 5 if( num.size() < 4 ) 6 { 7 return res; 8 } 9 vector<int> arr = num; 10 sort(arr.begin(), arr.end()); 11 for( int i = 0 ; i < arr.size()-3 ; i++ ) 12 { 13 if(i > 0 && arr[i] == arr[i-1]) 14 continue; 15 for( int j = i+1 ; j < arr.size()-2 ; j++ ) 16 { 17 if(j > i+1 && arr[j] == arr[j-1]) 18 continue; 19 int p = j+1; 20 int q = arr.size()-1; 21 while(p < q) 22 { 23 if (p > j + 1 && arr[p] == arr[p-1]) 24 { 25 p++; 26 continue; 27 } 28 if (q < arr.size() - 1 && arr[q] == arr[q+1]) 29 { 30 q--; 31 continue; 32 } 33 int sumtmp = arr[i] + arr[j] + arr[p] + arr[q] ; 34 if (sumtmp == target) 35 { 36 vector<int> tmp; 37 tmp.push_back(arr[i]); 38 tmp.push_back(arr[j]); 39 tmp.push_back(arr[p]); 40 tmp.push_back(arr[q]); 41 res.push_back(tmp); 42 p++; 43 } 44 else if( sumtmp < target ) 45 { 46 p++; 47 } 48 else 49 { 50 q--; 51 } 52 } 53 } 54 55 } 56 return res; 57 } 58 };