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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:
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)
和前面的思路基本上是相同的，这里同样容易遇见相同的数，这个时候需要跳过，代码如下：
1 class Solution { 2 public: 3 vector<vector<int>> fourSum(vector<int>& nums, int target) { 4 sort(nums.begin(), nums.end()); 5 int sz = nums.size(); 6 vector<vector<int>> ret; 7 for(int i = 0; i < sz; ++i){ 8 if(i != 0 && nums[i] == nums[i - 1]) 9 continue; 10 for(int j = i + 1; j < sz; ++j){ 11 if(j > i + 1 && nums[j] == nums[j - 1]) 12 continue; 13 for(int beg = j + 1, end = sz - 1; beg < end;){ 14 while(beg > j + 1 && nums[beg] == nums[beg - 1]) 15 beg++; 16 while(end < sz - 1 && nums[end] == nums[end + 1]) 17 end--; 18 if(beg >= end) break; 19 int sum = nums[i] + nums[j] + nums[beg] + nums[end]; 20 if(sum == target){ 21 vector<int> tmp; 22 tmp.push_back(nums[i]); 23 tmp.push_back(nums[j]); 24 tmp.push_back(nums[beg]); 25 tmp.push_back(nums[end]); 26 ret.push_back(tmp); 27 beg++, end--; 28 }else if(sum < target){ 29 beg++; 30 }else{ 31 end--; 32 } 33 } 34 } 35 } 36 return ret; 37 } 38 };
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4921928.html
