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]]

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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。

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Code（c++)

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> total;        int n = nums.size();        if(n<4)  return total;        sort(nums.begin(),nums.end());        for(int i=0;i<n-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[n-3]+nums[n-2]+nums[n-1]<target) continue;            for(int j=i+1;j<n-2;j++)            {                if(j>i+1&&nums[j]==nums[j-1]) continue;                if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;                if(nums[i]+nums[j]+nums[n-2]+nums[n-1]<target) continue;                int left=j+1,right=n-1;                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-1]&&left<right);                        do{right--;}while(nums[right]==nums[right+1]&&left<right);                    }                }            }        }        return total;     }};