【二分查找】自写二分函数的总结

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作者推荐

【动态规划】【广度优先搜索】LeetCode:2617 网格图中最少访问的格子数

本文涉及的基础知识点

二分查找算法合集

自写二分函数 的封装

我暂时只发现两种：
一，在左闭右开的区间寻找最后一个符合条件的元素，我封装成FindEnd函数。
二，在左开右闭的区间寻找第一个符合条件的元素，我封装成FindFirst函数。

namespace NBinarySearch
{template<class INDEX_TYPE,class _Pr>INDEX_TYPE FindFrist(INDEX_TYPE left, INDEX_TYPE rightInclue, _Pr pr){while (rightInclue - left > 1){const auto mid = left + (rightInclue - left) / 2;if (pr(mid)){rightInclue = mid;}else{left = mid;}}return rightInclue;}template<class INDEX_TYPE, class _Pr>INDEX_TYPE FindEnd(INDEX_TYPE leftInclude, INDEX_TYPE right, _Pr pr){while (right - leftInclude > 1){const auto mid = leftInclude + (right - leftInclude) / 2;if (pr(mid)){leftInclude = mid;}else{right = mid;}}return leftInclude;}
}

左闭右开 左开右闭的例子

LeetCode33: C++算法：二分查找旋转数组

class Solution {
public:int search(vector<int>& nums, int target) {int rBegin = NBinarySearch::FindFrist(-1, (int)nums.size() - 1, [&](const int mid) {return nums[mid] <= nums.back(); });if (target <= nums.back()){return Find(nums, rBegin, nums.size(), target);}return Find(nums, 0, rBegin, target);}int Find(const vector<int>& nums, int left, int r, int target){int iRet = NBinarySearch::FindEnd(left,r, [&](const int mid) {return nums[mid] <= target; });return (target == nums[iRet]) ? iRet : -1;}
};
int main()
{vector<int> vNums = { 1,2,3,4,6 };auto res = Solution().search(vNums, 4);std::cout << "index:" << res;if (-1 != res){std::cout << " value:" << vNums[res];}
}

左闭右开的例子

Leetcode2790:C++二分查找算法的应用：长度递增组的最大数目

namespace NBinarySearch
{template<class INDEX_TYPE,class _Pr>INDEX_TYPE FindFrist(INDEX_TYPE left, INDEX_TYPE rightInclue, _Pr pr){while (rightInclue - left > 1){const auto mid = left + (rightInclue - left) / 2;if (pr(mid)){rightInclue = mid;}else{left = mid;}}return rightInclue;}template<class INDEX_TYPE, class _Pr>INDEX_TYPE FindEnd(INDEX_TYPE leftInclude, INDEX_TYPE right, _Pr pr){while (right - leftInclude > 1){const auto mid = leftInclude + (right - leftInclude) / 2;if (pr(mid)){leftInclude = mid;}else{right = mid;}}return leftInclude;}
}class Solution {
public:int maxIncreasingGroups(vector<int>& usageLimits) {m_c = usageLimits.size();m_v = usageLimits;sort(m_v.begin(), m_v.end());auto Can = [&](int len){int i = m_c - 1;long long llNeed = 0;for (; len > 0; len--, i--){llNeed -= (m_v[i] - len);if (m_v[i] >= len){llNeed = max(0LL, llNeed);}}for (; i >= 0; i--){llNeed -= m_v[i];}return llNeed <= 0;};return NBinarySearch::FindEnd(1, m_c + 1, Can);}int m_c;vector<int> m_v;
};
template<class T>
void Assert(const T& t1, const T& t2)
{assert(t1 == t2);
}template<class T>
void Assert(const vector<T>& v1, const vector<T>& v2)
{if (v1.size() != v2.size()){assert(false);return;}for (int i = 0; i < v1.size(); i++){Assert(v1[i], v2[i]);}
}int main()
{Solution slu;vector<int> usageLimits;int res = 0;usageLimits = { 2,2,2 };res = slu.maxIncreasingGroups(usageLimits);Assert(res, 3);usageLimits = { 1,2,5 };res = slu.maxIncreasingGroups(usageLimits);Assert(res, 3);usageLimits = { 2,1,2 };res = slu.maxIncreasingGroups(usageLimits);Assert(res, 2);usageLimits = { 1,1 };res = slu.maxIncreasingGroups(usageLimits);Assert(res, 1);//CConsole::Out(res);
}

左闭右开的应用

C++二分查找算法的应用：最小好进制

Is

计算是否存在m位 k进制的1为目标数。m位iRet 进制1大于等于目标数，可能有多个符合的iRet，取最小值。非最小值一定不是。

namespace NBinarySearch
{template<class INDEX_TYPE,class _Pr>INDEX_TYPE FindFrist(INDEX_TYPE left, INDEX_TYPE rightInclue, _Pr pr){while (rightInclue - left > 1){const auto mid = left + (rightInclue - left) / 2;if (pr(mid)){rightInclue = mid;}else{left = mid;}}return rightInclue;}template<class INDEX_TYPE, class _Pr>INDEX_TYPE FindEnd(INDEX_TYPE leftInclude, INDEX_TYPE right, _Pr pr){while (right - leftInclude > 1){const auto mid = leftInclude + (right - leftInclude) / 2;if (pr(mid)){leftInclude = mid;}else{right = mid;}}return leftInclude;}
}class Solution {
public:string smallestGoodBase(string n) {long long llN = 0;for (const auto& ch : n){llN = (llN * 10 + ch - '0');}for (int i = 70; i > 2; i--){long long llRet = Is(i, llN);if (llRet > 0){return std::to_string(llRet);}}return std::to_string(llN - 1);}long long Is(int m, long long n){int iRet = NBinarySearch::FindEnd(2LL, n + 1LL, [&](const auto mid) {return Cmp(mid, m, n) >= 0; });return (0 == Cmp(iRet,m,n))? iRet : 0;}//k进制的m个1和n的大小比较,n大返回正数，相等返回0，n小返回负数long long Cmp(long long k, int m, long long n){long long llHas = 1;for (; m > 0; m--){if (n < llHas){return -1;}n -= llHas;if (m > 1){// 最后一次llHas并不使用，所以越界不影响if (LLONG_MAX / k < llHas){return -1;}llHas *= k;}}return n;}
};template<class T>
void Assert(const T& t1, const T& t2)
{assert(t1 == t2);
}template<class T>
void Assert(const vector<T>& v1, const vector<T>& v2)
{if (v1.size() != v2.size()){assert(false);return;}for (int i = 0; i < v1.size(); i++){Assert(v1[i], v2[i]);}
}int main()
{Solution slu;string res;res = slu.smallestGoodBase("470988884881403701");Assert(res, std::string("686286299"));res = slu.smallestGoodBase("2251799813685247");Assert(res, std::string("2"));res = slu.smallestGoodBase("13");Assert(res, std::string("3"));res = slu.smallestGoodBase("4681");Assert(res, std::string("8"));res = slu.smallestGoodBase("1000000000000000000");Assert(res, std::string("999999999999999999"));res = slu.smallestGoodBase("1333");Assert(res, std::string("36"));res = slu.smallestGoodBase("463381");Assert(res, std::string("463380"));//CConsole::Out(res);
}

扩展阅读

视频课程

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如何你想快

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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

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
如无特殊说明，本算法用**C++**实现。

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