LeetCode #18 4Sum

Description

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]
]
           

Analysis

题目难度为：Medium

本题思路遵循3Sum和2Sum问题，先去除一个数，剩下三个数才用3Sum来计算，也就是简化了3Sum问题。

Code（c++)

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> total;
        int n = nums.size();
        if(n<)  return total;
        sort(nums.begin(),nums.end());
        for(int i=;i<n-;i++)
        {
            if(i>&&nums[i]==nums[i-]) continue;
            if(nums[i]+nums[i+]+nums[i+]+nums[i+]>target) break;
            if(nums[i]+nums[n-]+nums[n-]+nums[n-]<target) continue;
            for(int j=i+;j<n-;j++)
            {
                if(j>i+&&nums[j]==nums[j-]) continue;
                if(nums[i]+nums[j]+nums[j+]+nums[j+]>target) break;
                if(nums[i]+nums[j]+nums[n-]+nums[n-]<target) continue;
                int left=j+,right=n-;
                while(left<right){
                    int sum=nums[left]+nums[right]+nums[i]+nums[j];
                    if(sum<target) left++;
                    else if(sum>target) right--;
                    else{
                        total.push_back(vector<int>{nums[i],nums[j],nums[left],nums[right]});
                        do{left++;}while(nums[left]==nums[left-]&&left<right);
                        do{right--;}while(nums[right]==nums[right+]&&left<right);
                    }
                }
            }
        }
        return total; 
    }
};