LeetCodet题解--18. 4Sum（4个数的和）
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 main
class 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