【算法】4 Sum 问题

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)

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> results;

        sort(nums.begin(), nums.end());

        int size = nums.size();
        for(int i = 0; i < size - 3; ++i) {
            if(i > 0 && nums[i] == nums[i-1]) continue;
            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;
            if(nums[i]+nums[size-3]+nums[size-2]+nums[size-1]<target) continue;

            vector<int> tmpVec(nums.begin() + i + 1, nums.end());
            threeSum(tmpVec, target - nums[i], results, nums[i]);
        }

        return results;
    }

    void threeSum(vector<int>& nums, int target, vector<vector<int>> &results, int d) {

        int size = nums.size();
        for(int i = 0; i < size - 2; ++i) {
            if ( i > 0 && (nums[i] == nums[i-1]) ) continue;
            int j = i + 1, k = size - 1;
            while( j < k) {

                while ( j > i + 1 && nums[j] == nums[j - 1]) {
                    ++j;
                }

                while ( k < size - 1 && nums[k] == nums[k + 1]) {
                    --k;
                    continue;
                }

                if (j >= k) break;

                int a = nums[i], b = nums[j], c = nums[k];

                int sum = a + b + c;
                if (sum == target) {
                    vector<int> result {d, a, b, c};
                    results.push_back(result);

                    ++j;
                    --k;
                }
                else if(sum < target){
                    ++j;
                }
                else {
                    --k;
                }

            }
        }
    }
};