[二分查找]LeetCode2040:两个有序数组的第 K 小乘积

• 算法结构

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& nums1,  vector& 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& nums1,  vector& 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& nums1, vector& 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 v12(it1, nums1.end());
		vector v22(it2, nums2.end());
		vector v11 = CopyAndMul(vector(nums1.begin(), it1));
		vector v21 = CopyAndMul(vector(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& nums11, vector& nums12,  vector& nums21,  vector& 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& nums1,  vector& 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 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;
	}
};

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;

};

扩展阅读

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我想对大家说的话
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如果程序是一条龙，那算法就是他的是睛

测试环境

操作系统：win7 开发环境： VS2019 C++17

或者 操作系统：win10 开发环境：

VS2022 C++17

[二分查找]LeetCode2040:两个有序数组的第 K 小乘积

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