和大神们学习每天一题(leetcode)-4Sum

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)

本题是在一个整数数组中查找和为目标值的四个元素，算法思想与2Sum和3Sum类似。
功能测试用例：{1,0,-1,0,-2,2},0
特殊测试用例：{},0;{0},0;{0,0},0;{0,0,0},0;{1,1,1,0,0,0,-1,-1,-1,0,-2,2,2,2},0
代码如下：

class Solution
{
public:
void FastSort(vector<int> &vsNumber, int nBegin, int nEnd)//快速排序
{
if (nBegin >= nEnd)
return;
int nPos1 = nBegin;
int sNMiddle = vsNumber[nEnd], sNChange;
for (int nPos2 = nBegin; nPos2 < nEnd; nPos2++)
{
if (vsNumber[nPos2] < sNMiddle)
{
sNChange = vsNumber[nPos1];
vsNumber[nPos1] = vsNumber[nPos2];
vsNumber[nPos2] = sNChange;
nPos1++;
}
}
vsNumber[nEnd] = vsNumber[nPos1];
vsNumber[nPos1] = sNMiddle;
FastSort(vsNumber, nBegin, nPos1 - 1);
FastSort(vsNumber, nPos1 + 1, nEnd);
}
vector<vector<int> > fourSum(vector<int> &num, int target)
{
vector<vector<int>> vvnResult;
vector<int> vnMiddle(4);
if (num.size() < 4)
return vvnResult;
FastSort(num,0,num.size()-1);//快速排序
int nBegin, nEnd, nRemain;
for (int nTemp1 = 0; nTemp1 < num.size() - 3; nTemp1++)//确定第一个数的位置
{
if (nTemp1 != 0 && num[nTemp1] == num[nTemp1 - 1])//去重复
continue;
for (int nTemp2 = nTemp1 + 1; nTemp2 < num.size() - 2; nTemp2++)//确定第二个数的位置
{
if (nTemp2 != nTemp1 + 1 && num[nTemp2] == num[nTemp2 - 1])//去重复
continue;
nBegin = nTemp2 + 1;
nEnd = num.size() - 1;
nRemain = target - num[nTemp1] - num[nTemp2];//简单前两个数后的剩余值
while (nBegin < nEnd)
{
if (nBegin != nTemp2 + 1 && num[nBegin] == num[nBegin - 1])//去重复
{
nBegin++;
}
else if (nEnd != num.size() - 1 && num[nEnd] == num[nEnd + 1])//去重复
{
nEnd--;
}
else//根据当前四个数的和确定后两个数的位置
{
if (num[nBegin] + num[nEnd] > nRemain)
{
nEnd--;
}
else if (num[nBegin] + num[nEnd] < nRemain)
{
nBegin++;
}
else
{
vnMiddle[0] = num[nTemp1];
vnMiddle[1] = num[nTemp2];
vnMiddle[2] = num[nBegin];
vnMiddle[3] = num[nEnd];
vvnResult.push_back(vnMiddle);
nEnd--;
nBegin++;
}
}
}
}
}
return vvnResult;
}
};