LeetCodet题解--18. 4Sum（4个数的和）

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链接

LeetCode题目：https://leetcode.com/problems/4sum

GitHub代码：https://github.com/gatieme/LeetCode/tree/master/018-4Sum

CSDN题解：http://blog.csdn.net/gatieme/article/details/51089460

题意

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)

给一个包含n个数的整数数组S，在S中找到所有使得和为给定整数target的四元组(a, b, c, d)。
例如，对于给定的整数数组S=[1, 0, -1, 0, -2, 2]和target=0. 满足要求的四元组集合为：

(-1, 0, 0, 1)(-2, -1, 1, 2)(-2, 0, 0, 2)

四元组(a, b, c, d)中，需要满足a <= b <= c <= d
答案中不可以包含重复的四元组

借用3Sum的方法

我们采用2sum，以及3sum的方法，先排序后，用两趟循环循环分别遍历第一个数和第二个数，然后剩余的两个数，用二分查找的方法去找。

对数组排序
确定四元数中的前两个（a，b）
遍历剩余数组确定两外两个（c，d），确定cd时思路跟3Sum确定后两个数据一样，二分查找左右逼近。

#include <iostream>#include <vector>#include <algorithm>using namespace std;//#define __tmain mainclass Solution{public:    vector<vector<int>> fourSum(vector<int>& nums, int target)    {        vector<vector<int>> res;        int size = nums.size( );        sort(nums.begin( ), nums.end( ));        for(int i = 0; i < size - 3; i++)        {            //skip same i            while(i > 0 && i < size - 3 && nums[i] == nums[i - 1])            {                //cout <<"skip all " <<nums[i] <<endl;                i++;            }            for(int j = i + 1; j < size - 2; j++)            {                //skip same i                while(j > i + 1 && j < size - 2 && nums[j] == nums[j - 1])                {                    //cout <<"skip all " <<nums[i] <<endl;                    j++;                }                for(int start = j + 1, end = size - 1;                    start < end; )                {                    int sum = nums[i] + nums[j] + nums[start] + nums[end];                    if(sum == target)                    {                        vector<int> cur(4);                        cur[0] = nums[i];                        cur[1] = nums[j];                        cur[2] = nums[start];                        cur[3] = nums[end];                        res.push_back(cur);                        start++;                        end--;                        //skip same j                        while(start < end                           && nums[start] == nums[start - 1])                        {                            //cout <<"skip all " <<nums[start] <<endl;                            start++;                        }                        //skip same k                        while(end > start && nums[end] == nums[end + 1])                        {                            //cout <<"skip all " <<nums[end] <<endl;                            end--;                        }                    }                    else if(sum > target)                    {                        end--;                        while(end > start && nums[end] == nums[end + 1])                        {                            //cout <<"skip all " <<nums[end] <<endl;                            end--;                        }                    }                    else if(sum < target)                    {                        start++;                        while(start < end                           && nums[start] == nums[start - 1])                        {                            //cout <<"skip all " <<nums[start] <<endl;                            start++;                        }                    }                }            }        }        return res;    }};int __tmain(){    vector<vector<int>> result;    Solution solution;    vector<int> vec;    vec.push_back(-3);    vec.push_back(-2);    vec.push_back(-1);    vec.push_back(0);    vec.push_back(0);    vec.push_back(1);    vec.push_back(2);    vec.push_back(3);    result = solution.fourSum(vec, 0);    for(int i = 0; i < result.size( ); i++)    {        for(int j = 0; j < result[i].size( ); j++)        {            printf("%d ", result[i][j]);        }        printf("\n");    }    return 0;}

我们是通过跳过相同的值的方法来去重复的

            //skip same i            while(i > 0 && i < size - 3 && nums[i] == nums[i - 1])            {                //cout <<"skip all " <<nums[i] <<endl;                i++;            }

网上同样思路的方法，看到有些是通过set来去重复的

参见 [LeetCode]18.4Sum

#include <iostream>#include <stdio.h>#include <vector>#include <set>#include <algorithm>using namespace std;class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target)    {        int i,j,start,end;        int Len = num.size();        vector<int> triplet;        vector<vector<int>> triplets;        set<vector<int>> sets;        //排序        sort(num.begin(),num.end());        for(i = 0;i < Len-3;i++)        {            for(j = i + 1;j < Len - 2;j++)            {                //二分查找                start = j + 1;                end = Len - 1;                while(start < end){                    int curSum = num[i] + num[j] + num[start] + num[end];                    //相等 -> 目标                    if(target == curSum)                    {                        triplet.clear();                        triplet.push_back(num[i]);                        triplet.push_back(num[j]);                        triplet.push_back(num[start]);                        triplet.push_back(num[end]);                        sets.insert(triplet);                        start ++;                        end --;                    }                    //大于 -> 当前值小需要增大                    else if(target > curSum)                    {                        start ++;                    }                    //小于 -> 当前值大需要减小                    else{                        end --;                    }                }//while            }        }//for        //利用set去重        set<vector<int>>::iterator it = sets.begin();        for(; it != sets.end(); it++)            triplets.push_back(*it);        return triplets;    }};int main() {    vector<vector<int>> result;    Solution solution;    vector<int> vec;    vec.push_back(-3);    vec.push_back(-2);    vec.push_back(-1);    vec.push_back(0);    vec.push_back(0);    vec.push_back(1);    vec.push_back(2);    vec.push_back(3);    result = solution.fourSum(vec,0);    for(int i = 0;i < result.size();i++)    {        for(int j = 0;j < result[i].size();j++)        {            cout <<result[i][j]);        }        cout <<endl;    }    return 0;}

递归求解

类似于15题求解的3Sum问题，这次求解4Sum问题，本质是相同的，不可以采用穷举法；
其实求解4Sum问题可以分解为求3Sum问题，对数列依次遍历i，我们只需得到在第i个数后面，找出所有和为（target?nums[i]）的三元组，同理求3Sum又可以退化为2Sum，进而退化为1Sum。
因此，采用递归的思想解决KSum问题。

#include <iostream>#include <cstdlib>#include <vector>#include <algorithm>#include <unordered_set>#include <set>using namespace std;class Solution {public:    /*4-sum算法，递归实现，TLE*/    vector<vector<int>> fourSum1(vector<int>& nums, int target) {        if (nums.empty())            return vector<vector<int>>();        sort(nums.begin(), nums.end());        return k_Sum(nums, 0, 4, target);    }    /*k-sum算法*/    vector<vector<int>> k_Sum(vector<int> &nums, int begPos, int count, int target)    {        if (nums.empty())            return vector<vector<int>>();        /*所输入序列为已排序*/        int len = nums.size();        unordered_set<int> visited;        vector<vector<int>> ret;        vector<int> tmp;        /*2-sum 处理*/        if (2 == count)        {            int i = begPos, j = len - 1;            while (i < j)            {                int sum = nums[i] + nums[j];                if (sum == target && visited.find(nums[i]) == visited.end())                {                    tmp.clear();                    tmp.push_back(nums[i]);                    tmp.push_back(nums[j]);                    ret.push_back(tmp);                    /*加入已访问set*/                    visited.insert(nums[i]);                    visited.insert(nums[j]);                    ++i;                    --j;                }//if                else if (sum < target)                    ++i;                else                    --j;            }//while        }//if        else{            for (int i = begPos; i < len; ++i)            {                if (visited.find(nums[i]) == visited.end())                {                    visited.insert(nums[i]);                    /*得到k-1 sum的序列*/                    vector<vector<int>> subRet = k_Sum(nums, i+1, count - 1, target-nums[i]);                    if (!subRet.empty())                    {                        int sz = subRet.size();                        for (int j = 0; j < sz; ++j)                        {                            subRet[j].insert(subRet[j].begin(), nums[i]);                        }//for                        ret.insert(ret.end(), subRet.begin(), subRet.end());                    }//if                }//if            }//for        }//else        /*返回结果集*/        return ret;    }    /*4-sum算法，方法二，2-sum的变形*/    vector<vector<int>> fourSum(vector<int>& nums, int target) {        if (nums.empty() || nums.size() < 4)            return vector<vector<int>>();        sort(nums.begin(), nums.end());        int len = nums.size();        set<vector<int>> tmpRet;        vector<vector<int>> res;        for (int i = 0; i < len; ++i)        {            for (int j = i + 1; j < len; ++j)            {                int beg = j + 1, end = len - 1;                while (beg < end)                {                    int sum = nums[i] + nums[j] + nums[beg] + nums[end];                    if (sum == target)                    {                        vector<int> tmp = { nums[i], nums[j], nums[beg], nums[end] };                        tmpRet.insert(tmp);                        ++beg;                        --end;                    }                    else if (sum < target)                    {                        ++beg;                    }                    else                        --end;                }//while            }//for        }//for        auto iter = tmpRet.begin();        while (iter != tmpRet.end())        {            res.push_back(*iter);            ++iter;        }//while        return res;    }};int main(){    Solution s;    vector<int> v = { 1, 0, -1, 0, -2, 2 };    vector<vector<int>>ret = s.fourSum(v,0);    system("pause");    return

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