[二分查找]LeetCode2040:两个有序数组的第 K 小乘积

本文涉及的基础知识点

二分查找算法合集

题目

给你两个从小到大排好序且下标从 0 开始的整数数组 nums1 和 nums2 以及一个整数 k ，请你返回第 k （从 1 开始编号）小的 nums1[i] * nums2[j] 的乘积，其中 0 <= i < nums1.length 且 0 <= j < nums2.length 。
示例 1：
输入：nums1 = [2,5], nums2 = [3,4], k = 2
输出：8
解释：第 2 小的乘积计算如下：

nums1[0] * nums2[0] = 2 * 3 = 6
nums1[0] * nums2[1] = 2 * 4 = 8
第 2 小的乘积为 8 。
示例 2：
输入：nums1 = [-4,-2,0,3], nums2 = [2,4], k = 6
输出：0
解释：第 6 小的乘积计算如下：
nums1[0] * nums2[1] = (-4) * 4 = -16
nums1[0] * nums2[0] = (-4) * 2 = -8
nums1[1] * nums2[1] = (-2) * 4 = -8
nums1[1] * nums2[0] = (-2) * 2 = -4
nums1[2] * nums2[0] = 0 * 2 = 0
nums1[2] * nums2[1] = 0 * 4 = 0
第 6 小的乘积为 0 。
示例 3：
输入：nums1 = [-2,-1,0,1,2], nums2 = [-3,-1,2,4,5], k = 3
输出：-6
解释：第 3 小的乘积计算如下：
nums1[0] * nums2[4] = (-2) * 5 = -10
nums1[0] * nums2[3] = (-2) * 4 = -8
nums1[4] * nums2[0] = 2 * (-3) = -6
第 3 小的乘积为 -6 。
参数范围：
1 <= nums1.length, nums2.length <= 5 * 10⁴
-10⁵ <= nums1[i], nums2[j] <= 10⁵
1 <= k <= nums1.length * nums2.length
nums1 和 nums2 都是从小到大排好序的。

两层二分查找

时间复杂度

O(log(max²)nlogn),n是两个数组长度的较大者，max 是两个数组的最大值。

分情况讨论

结果	数组一	数组二
负数	负数	正数
负数	正数	负数
0	0	任意数
0	非0	0
正数	正数	正数
正数	负数	负数

第一层二分

寻找一个符合如下条件的llMul：
乘积小于等于llMul的组合数量大于等于k。
左开右闭空间。

负数的问题

如果乘积为负数，第k小则绝对值第k大。我们可以负数全部转成绝对值，然后倒序，这样可以保证升序。m个数，第k大（从1开始），就是m-k+1小。

变量解释


v11	数组一中的负数的绝对值，升序
v12	数组一中的正数，升序
v21	数组二中的负数的绝对值，升序
v22	数组二中的正数，升序

代码

核心代码

class Solution {
public:
long long kthSmallestProduct(vector& nums1, vector& nums2, long long k) {
auto it1 = std::equal_range(nums1.begin(), nums1.end(), 0);
auto it2 = std::equal_range(nums2.begin(), nums2.end(), 0);
const long long less0Count1 = it1.first - nums1.begin();
const long long i0Count1 = it1.second - it1.first;
const long long great0Count1 = nums1.end() - it1.second;
const long long less0Count2 = it2.first - nums2.begin();
const long long i0Count2 = it2.second - it2.first;
const long long great0Count2 = nums2.end() - it2.second;
const long long llZeroCount = i0Count1 * nums2.size() + i0Count2 * nums1.size() - i0Count1 * i0Count2;
const long long llLess0Cout = less0Count1 * great0Count2 + less0Count2 * great0Count1;
vector v12(it1.second, nums1.end());
vector v22(it2.second, nums2.end());
vector v11 = CopyAndMul(vector(nums1.begin(), it1.first));
vector v21 = CopyAndMul(vector(nums2.begin(), it2.first));
if (k <= llLess0Cout)
{//在负数中找
k = llLess0Cout + 1 - k;
return -DoGreate0(v11, v22, v21, v12, k);
}
k -= llLess0Cout;
if (k <= llZeroCount)
{
return 0;
}
k -= llZeroCount;
return DoGreate0(v11, v21,v12, v22,k);
}
//从升序正数数组中寻找第k小的积：第一个积小于等于llMul 的数量大于等于k 左开右闭
long long DoGreate0(const vector& nums11,const vector& nums12, const vector& nums21, const vector& nums22, long long k)
{
long long left = 0, right = (long long) 1e10;
while (right - left > 1)
{
const auto mid = left + (right - left) / 2;
int iCnt = 0;
const long long llHas = LessEqual(nums11, nums12, mid) + LessEqual(nums21, nums22, mid);
if (llHas >= k)
{
right = mid;
}
else
{
left = mid;
}
}
return right;
}
long long LessEqual(const vector& nums1, const vector& nums2, long long llMul)
{
long long llCnt = 0;
for (const auto& n : nums2)
{
llCnt += std::upper_bound(nums1.begin(), nums1.end(), llMul / n) - nums1.begin();
}
return llCnt;
}
vector CopyAndMul(const vector& nums)
{
vector vRet(nums.size());
for (int i = 0; i < nums.size(); i++)
{
vRet[i] = -nums[nums.size() - 1 - i];
}
return vRet;
}
};

测试用例

template
void Assert(const vector& v1, const vector& v2)
{
if (v1.size() != v2.size())
{
assert(false);
return;
}
for (int i = 0; i < v1.size(); i++)
{
assert(v1[i] == v2[i]);
}
}

template
void Assert(const T& t1, const T& t2)
{
assert(t1 == t2);
}

int main()
{
vector nums1, nums2;
long long k, res;
{
nums1 = { -2,-1 }, nums2 = { -2,-1 }, k = 4;
Solution slu;
auto res = slu.kthSmallestProduct(nums1, nums2, k);
Assert(4LL, res);
}
{
nums1 = { 2, 5 }, nums2 = { 3, 4 }, k = 2;
Solution slu;
auto res = slu.kthSmallestProduct(nums1, nums2, k);
Assert(8LL, res);
}
{
nums1 = { -4,-2,0,3 }, nums2 = { 2,4 }, k = 6;
Solution slu;
auto res = slu.kthSmallestProduct(nums1, nums2, k);
Assert(0LL, res);
}
{
nums1 = { -2,-1,0,1,2 }, nums2 = { -3,-1,2,4,5 }, k = 3;
Solution slu;
auto res = slu.kthSmallestProduct(nums1, nums2, k);
Assert(-6LL, res);
}
{
nums1 = { 0 }, nums2 = { 0,0,0}, k = 3;
Solution slu;
auto res = slu.kthSmallestProduct(nums1, nums2, k);
Assert(0LL, res);
}
{
nums1 = { 1,2 }, nums2 = { 1,2}, k = 3;
Solution slu;
auto res = slu.kthSmallestProduct(nums1, nums2, k);
Assert(2LL, res);
}
{
nums1 = { 1,10000 };
nums2 = { 1,10000 };
k = 4;
Solution slu;
auto res = slu.kthSmallestProduct(nums1, nums2, k);
Assert(10000* 10000LL, res);
}

//CConsole::Out(res);

}

优化一

确保数组二的长度比数组一短

if (nums1.size() < nums2.size())
		{
			swap(nums1, nums2);
		}

完整函数：

	long long LessEqual( vector<int>& nums1,  vector<int>& nums2, long long llMul)
	{
		if (nums1.size() < nums2.size())
		{
			swap(nums1, nums2);
		}
		long long llCnt = 0;
		for (const auto& n : nums2)
		{
			llCnt += std::upper_bound(nums1.begin(), nums1.end(), llMul / n) - nums1.begin();
		}
		return llCnt;
	}

优化二

第二层二分查找可以优化成双指针。这样也不用思考取整之类，容易理解。

	long long LessEqual( vector<int>& nums1,  vector<int>& nums2, long long llMul)
	{
		long long llCnt = 0;
		int right = nums1.size()-1;
		for (const auto& n : nums2)
		{
			while ((right >=0 ) && (nums1[right] * (long long)n > llMul))
			{//nums1[0,right]*n 全部小于等于llMul
				right--;
			}
			llCnt += (right+1);
		}
		return llCnt;
	}

优化三

0不必单独考虑。0符合负数的规则：绝对值越大，乘积越小。0也符合正数的规则，觉得值越大，乘积越大。

class Solution {
public:
	long long kthSmallestProduct(vector<int>& nums1, vector<int>& nums2, long long k) {
		auto it1 = std::lower_bound(nums1.begin(), nums1.end(), 0);
		auto it2 = std::lower_bound(nums2.begin(), nums2.end(), 0);
		vector<int> v12(it1, nums1.end());
		vector<int> v22(it2, nums2.end());
		vector<int> v11 = CopyAndMul(vector<int>(nums1.begin(), it1));
		vector<int> v21 = CopyAndMul(vector<int>(nums2.begin(), it2));
		const long long ll24Count = v11.size() * (long long)v22.size() + (long long)v12.size() * v21.size();
		if (k <= ll24Count)
		{//在负数中找
			k = ll24Count + 1 - k;
			return -DoGreate0(v11, v22, v21, v12, k);
		}
		k -= ll24Count;	
		return DoGreate0(v11, v21,v12, v22,k);
	}
	//从升序正数数组中寻找第k小的积： 第一个积小于等于llMul 的数量大于等于k 左开右闭
	long long DoGreate0( vector<int>& nums11, vector<int>& nums12,  vector<int>& nums21,  vector<int>& nums22, long long k)
	{
		long long left = -1, right = (long long) 1e10;
		while (right - left > 1)
		{
			const auto mid = left + (right - left) / 2;
			const long long llHas = LessEqual(nums11, nums12, mid) + LessEqual(nums21, nums22, mid);
			if (llHas >= k)
			{
				right = mid;
			}
			else
			{
				left = mid;
			}
		}
		return right;
	}
	long long LessEqual( vector<int>& nums1,  vector<int>& nums2, long long llMul)
	{
		long long llCnt = 0;
		int right = nums1.size()-1;
		for (const auto& n : nums2)
		{
			while ((right >=0 ) && (nums1[right] * (long long)n > llMul))
			{//nums1[0,right]*n 全部小于等于llMul
				right--;
			}
			llCnt += (right+1);
		}
		return llCnt;
	}
	vector<int> CopyAndMul(const vector<int>& nums)
	{
		vector<int> vRet(nums.size());
		for (int i = 0; i < nums.size(); i++)
		{
			vRet[i] = -nums[nums.size() - 1 - i];
		}
		return vRet;
	}
};

2023年3月版

class CNumHelp
{
public:
CNumHelp(vector& nums) :m_nums(nums)
{
auto it1 = std::equal_range(m_nums.begin(), m_nums.end(), 0);
m_iLess0Num = it1.first - m_nums.begin();
m_i0Num = it1.second - it1.first;
m_iMore0Num = m_nums.end() - it1.second;
m_iLessEqual0Num = m_iLess0Num + m_i0Num;
m_iMoreEqualNum = m_iMore0Num + m_i0Num;
}
vector m_nums;
int m_iLess0Num = 0, m_i0Num = 0, m_iMore0Num = 0;
int m_iLessEqual0Num = 0,m_iMoreEqualNum=0;
};

class ICal
{
public:
virtual long long Cal(long long llMid)const = 0;
};
class CCalMore0 : public ICal
{
public:
CCalMore0(const CNumHelp& help1, const CNumHelp& help2) :m_help1(help1), m_help2(help2)
{

 }
 virtual long long Cal(long long llMid)const
 {
	 long long llNum = 0;
	 for (int i = m_help1.m_iLessEqual0Num; i < m_help1.m_nums.size(); i++)
	 {
		 int iCurNum = std::upper_bound(m_help2.m_nums.begin(), m_help2.m_nums.end(), llMid / m_help1.m_nums[i]) - m_help2.m_nums.begin() - m_help2.m_iLessEqual0Num;
		 llNum += iCurNum;
	 }

	 for (int i = 0; i < m_help1.m_iLess0Num; i++)
	 {
		 auto it = std::equal_range(m_help2.m_nums.begin(), m_help2.m_nums.end(), llMid / m_help1.m_nums[i]);
		 //auto it2 = (0 == llMid % m_help1.m_nums[i]) ? it.first : it.second;
		 auto it2 = it.first;
		 llNum += m_help2.m_nums.end() - it2 - m_help2.m_iMoreEqualNum;
	 }
	 return llNum;
 }

private:

 const CNumHelp m_help1;
 const CNumHelp m_help2;

};

class CCalLess0 : public ICal
{
public:
CCalLess0(const CNumHelp& help1, const CNumHelp& help2) :m_help1(help1), m_help2(help2)
{
}
virtual long long Cal(long long llMid)const
{
return Cal(llMid, m_help1, m_help2) + Cal(llMid, m_help2, m_help1);
}
static long long Cal(long long llMid, const CNumHelp& help1, const CNumHelp& help2)
{
long long llNum = 0;
for (int i = help1.m_iLessEqual0Num; i < help1.m_nums.size(); i++)
{
auto it = std::equal_range(help2.m_nums.begin(), help2.m_nums.end(), llMid / help1.m_nums[i]);
auto it2 = (0 == llMid% help1.m_nums[i]) ? it.second : it.first;
int iCurNum = it2 -help2.m_nums.begin();
llNum += iCurNum;
}
return llNum;
}
private:
const CNumHelp m_help1;
const CNumHelp m_help2;
};
class Solution {
public:
long long kthSmallestProduct(vector& nums1, vector& nums2, long long k) {
CNumHelp help1(nums1), help2(nums2);
//const long long llTotal = (long long)nums1.size()nums2.size();
const long long ll0Num = (long long)help1.m_i0Num * nums2.size() + (long long)help2.m_i0Num * nums1.size() - (long long)help1.m_i0Numhelp2.m_i0Num;
const long long llLess0Num = (long long)help1.m_iMore0Num * help2.m_iLess0Num + (long long)help1.m_iLess0Num * help2.m_iMore0Num;
if (k <= llLess0Num)
{
CCalLess0 cal(help1, help2);
return Do(cal, k, (long long)100000 * -100000 - 1,-1);
}
k -= llLess0Num;
if (k <= ll0Num)
{
return 0;
}
k -= ll0Num;
CCalMore0 cal(help1, help2);
return Do(cal, k, 0, (long long)100000 * 100000);
}
long long Do(const ICal& cal, long long k, long long left, long right)
{
while (right > left + 1)
{
const auto llMid = left + (right - left) / 2;
const long long llNum = cal.Cal(llMid);
if (llNum >= k)
{
right = llMid;
}
else
{
left = llMid;
}
}
return right;
}

};

2023年9月

class Solution {
public:
long long kthSmallestProduct(const vector& nums1, const vector& nums2, long long k) {
CalRange(nums1, m_v11, m_v12);
CalRange(nums2, m_v21, m_v22);
const int iZero1Num = nums1.size() - m_v11.size() - m_v12.size();
const int iZero2Num = nums2.size() - m_v21.size() - m_v22.size();
long long llLess0 = (long long)m_v11.size() * m_v22.size() + (long long)m_v21.size() * m_v12.size();
long long ll0 = (long long)nums2.size() * iZero1Num + (long long)nums1.size() * iZero2Num - (long long)iZero1Num * iZero2Num;
if (k <= llLess0)
{//结果是负数
m_v21.swap(m_v22);
return -Do(llLess0 - k + 1);
}
k -= llLess0;
if (k <= ll0)
{
return 0;
}
k -= ll0;
return Do(k);
}
long long Do(long long k)
{
long long left =-(1e10 + 0.5)-1, r = 1e10 + 0.5;
while (r - left > 1)
{
const auto mid = left + (r - left) / 2;
long long llNum = CountEqualLess(m_v11, m_v21,mid) + CountEqualLess(m_v12, m_v22,mid);
if (llNum >= k)
{
r = mid;
}
else
{
left = mid;
}
}
return r;
}
long long CountEqualLess(const vector& nums1, const vector& nums2, long long llMul)
{
long long llCnt = 0;
int r = 0 ;//[0,r)和num2[i]的乘积 < llMul
for (int i =nums2.size()-1; i >= 0 ;i-- )
{
for (; (r < nums1.size() ) && ((long long)nums1[r] * nums2[i] <= llMul); r++);
llCnt += r;
}
return llCnt;
}
static void CalRange(const vector& nums, vector& v1, vector& v2)
{
int i = 0;
for (i = 0; (i < nums.size()) && (nums[i] < 0); i++)
{
v1.emplace_back(-nums[i]);
}
std::reverse(v1.begin(), v1.end());
for (; (i < nums.size()) && (nums[i] == 0); i++);
for (; i < nums.size(); i++)
{
v2.emplace_back(nums[i]);
}
}
vector m_v11, m_v12, m_v21, m_v22;

};

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我想对大家说的话
闻缺陷则喜是一个美好的愿望，早发现问题，早修改问题，给老板节约钱。
子墨子言之：事无终始，无务多业。也就是我们常说的专业的人做专业的事。
如果程序是一条龙，那算法就是他的是睛