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.

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]

]

思路：

类似3sum求和，然后在过程中及时约束，判断是否可以提前跳出循环来优化。

vector
     
      
       > fourSum(vector
       
        & nums, int target)    {        vector
        
         
          >result;        const int size = nums.size();        if (size < 4) return result;        sort(nums.begin(), nums.end());        vector
          
           temp; for (int i = 0; i < size - 3; i++) { if (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) break;//及时跳出循环 if (nums[i] + nums[size - 3] + nums[size - 2] + nums[size - 1] < target)continue; for (int j = i + 1; j < size-2; j++) { if (nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) break;//及时跳出循环 if (nums[i] + nums[j] + nums[size - 2] + nums[size - 1] < target)continue; int left = j + 1; int right = size - 1; while (left < right) { int sum = nums[i] + nums[j] + nums[left] + nums[right]; if (sum < target)left++; else if (sum > target) right--; else { temp = {nums[i],nums[j],nums[left],nums[right]};//新学一招，类似数组赋值 result.push_back(temp); while (left < right && nums[left] == temp[2])left++; while (left < right && nums[right] == temp[3])right--; } } while (j + 1 < size && nums[j + 1] == nums[j])j++; } while (i + 1 < size && nums[i + 1] == nums[i])i++; } return result; }
          
         
        
       
      
     

参考：