LeetCode 18. 4Sum

题目描述

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: 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]]

题目解析

思路一：
先利用map特性排序，以第i个为头，从右边找是否存在3sum。在3sum中同样使用2sum。
但是超时了,但是思路应该是对的，时间复杂度为O(n^3)。

思路二：
看了题目的标签有 twopointer，将2sum过程改用twopointer方法。使用两个循环；

% 1. 避免重复if(i>0&&nums[i]==nums[i-1])continue;% 2. 在循环开始时进行检查if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;if(nums[i]+nums[cnt-3]+nums[cnt-2]+nums[cnt-1]<target) continue;

代码（思路一）

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int> >ans_set;        vector<int> ans(4,0);        map<int,int>nums_map;        for(int it:nums)        {            nums_map[it]++;        }        for(auto it = nums_map.begin();it!=nums_map.end();it++)        {            ans[0]=it->first;            int threesum = target - it->first;            find_3sum(ans_set,ans,nums_map,threesum);        }        return ans_set;    }    void find_3sum(vector<vector<int> >&ans_set,vector<int> ans,map<int,int>&nums_map,int threesum)    {        nums_map[ans[0]]--;        for(auto it=nums_map.find(ans[0]);it!=nums_map.end();it++)        {            if(it->second<=0)            {                continue;            }            ans[1]=it->first;            int twosum = threesum - it->first;            find_2sum(ans_set,ans,nums_map,twosum);        }        nums_map[ans[0]]++;        return;    }    void find_2sum(vector<vector<int> >&ans_set,vector<int> ans,map<int,int>&nums_map,int twosum)    {        nums_map[ans[1]]--;        for(auto it=nums_map.find(ans[1]);it!=nums_map.end();it++)        {            if(it->second<=0)            {                continue;            }            int p = twosum - it->first;            if(nums_map.count(p)>0)            {                if(p>it->first||p==it->first&&it->second>1)                {                    ans[2] = it->first;                    ans[3] = p;                    ans_set.push_back(ans);                }            }        }        nums_map[ans[1]]++;        return;    }};

代码（思路二）

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        sort(nums.begin(),nums.end());        vector<vector<int> >ans_set;        vector<int> ans(4,0);        int cnt = nums.size();        for(int i = 0;i < cnt-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[cnt-3]+nums[cnt-2]+nums[cnt-1]<target) continue;            for(int j = i+1;j < cnt-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[cnt-2]+nums[cnt-1]<target) continue;                int p = j+1;                int q = cnt-1;                while(p<q)                {                    if(nums[i]+nums[j]+nums[p]+nums[q]>target)                    {                        q--;                        while(nums[q]==nums[q+1]&&p<q)                        {                            q--;                        }                    }                    else if(nums[i]+nums[j]+nums[p]+nums[q]<target)                    {                        p++;                        while(nums[p]==nums[p-1]&&p<q)                        {                            p++;                        }                    }                    else                    {                        ans[0]=nums[i];                        ans[1]=nums[j];                        ans[2]=nums[p];                        ans[3]=nums[q];                        ans_set.push_back(ans);                        q--;                        while(nums[q]==nums[q+1]&&p<q)                        {                            q--;                        }                        p++;                        while(nums[p]==nums[p-1]&&p<q)                        {                            p++;                        }                    }                }            }        }        return ans_set;    }};