leetcode18—4Sum

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

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [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]
]

想法：类似于3Sum的解决方式，设立双指针求解

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int len = nums.size();
        vector<vector<int>> result;
         == len)
            return result;
        sort(nums.begin(),nums.end());
         ; i < len- ; i++){
             && nums.at(i) == nums.at(i-))
                continue;
             && nums.at(i) + nums.at(i+)+nums.at(i+)+nums.at(i+) > target)
                break;
             && nums.at(i) + nums.at(len-)+nums.at(len-)+nums.at(len-) < target)
                continue;
                 ; j < len- ; j++){
                     && nums[j] == nums[j-])
                        continue;
                     && nums[i] + nums[j] + nums[j+] + nums[j+] > target)
                        break;
                     && nums[i] + nums[j] + nums[len-] + nums[len-] < target)
                        continue;

                    ;
                        ;
                        while(left < right){
                            int temp = nums.at(i) + nums.at(j) + nums.at(left) + nums.at(right);
                            if(temp == target){
                                result.push_back({nums.at(i),nums.at(j),nums.at(left),nums.at(right)});
                                left++;
                                right--;
                                ])//避免重复
                                    left++;
                                ])
                                    right--;
                            }else if(temp < target){
                                left++;
                            }else{
                                right--;
                            }
                        }

                }

        }
        return result;

    }
};