leetcode18. 4Sum 题解

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]
]
题目大意：给一个数组和target，求数组中所有使a+b+c+d = target的可能。
题目思路：和3sum思路类似，构建4个索引，时间复杂度为O(n³).
代码如下：

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int l,m,n,k,sum;
        int len = nums.size();
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        for(int l = 0; l < len-3; l++){
            if(nums[l]+nums[l+1]+nums[l+2]+nums[l+3] > target)
                break;
            if(nums[len-4]+nums[len-3]+nums[len-2]+nums[len-1] < target)
                break;
            for(m = l+1; m < len-2; m++){
                int start = m+1, end = len-1;
           //     cout<<"start = "<<start<<" "<<"end = "<<end<<endl;
                while(start < end){
                    sum = nums[l] + nums[m] + nums[start] + nums[end];

                    if(sum < target){
                        ++start;

                    }
                    if(sum > target){
                        --end;
                    }
                    if(sum == target){
                //        cout<<"start = "<<start<<endl;
                //        cout<<"end = "<<end<<endl;
                        ans.push_back({nums[l] , nums[m] , nums[start] , nums[end]});
                        ++start;
                        while(nums[start] == nums[start-1] && start < end)
                            ++start;
                        while(nums[end] == nums[end+1] && start < end)
                            --end;
                    }
                }

                while(m < len-2 && nums[m] == nums[m+1])
                        m++;
             //  cout<<"now m = "<<m<<endl;

             /*   */
            }
            while(l < len-3 && nums[l] == nums[l+1])
                    ++l;

        }
        return ans;
    }
};