[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 = 6nums1[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 = -16nums1[0] * nums2[0] = (-4) * 2 = -8nums1[1] * nums2[1] = (-2) * 4 = -8nums1[1] * nums2[0] = (-2) * 2 = -4nums1[2] * nums2[0] = 0 * 2 = 0nums1[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 = -10nums1[0] * nums2[3] = (-2) * 4 = -8nums1[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00任意数0非00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
视频课程
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如何你想快
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相关下载
想高屋建瓴的学习算法，请下载《喜缺全书算法册》doc版 https://download..net/download/he_zhidan/88348653
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
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
