【二分查找】自写二分函数的总结
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【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
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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。
