二进制搜索是竞争性编程或任何算法竞争中的重要组成部分，因为具有速记功能的知识，可以减少对它们进行编码的时间。该搜索仅在对容器进行排序时有效。相关功能在下面讨论。
1.binary_search(start_ptr，end_ptr，num) ：如果元素存在于容器中，则此函数返回boolean true ，否则返回false。
// C++ code to demonstrate the working of binary_search()
#include
using namespace std;
int main()
{
// initializing vector of integers
vector arr = {10, 15, 20, 25, 30, 35};
// using binary_search to check if 15 exists
if (binary_search(arr.begin(), arr.end(), 15))
cout << "15 exists in vector";
else
cout << "15 does not exist";
cout << endl;
// using binary_search to check if 23 exists
if (binary_search(arr.begin(), arr.end(), 23))
cout << "23 exists in vector";
else
cout << "23 does not exist";
cout << endl;
}
输出：
15 exists in vector
23 does not exist
2.lower_bound(start_ptr，end_ptr，num) ：如果容器包含1次num，则返回指向“ num的位置”的指针。如果容器包含num多次出现，则返回指向“ num的第一个位置”的指针。如果容器不包含num，则返回指向“下一个比num高的数字的位置”的指针。将指针减去到第一位置即“ vect.begin() ”将返回实际索引。
// C++ code to demonstrate the working of lower_bound()
#include
using namespace std;
int main()
{
// initializing vector of integers
// for single occurrence
vector arr1 = {10, 15, 20, 25, 30, 35};
// initializing vector of integers
// for multiple occurrences
vector arr2 = {10, 15, 20, 20, 25, 30, 35};
// initializing vector of integers
// for no occurrence
vector arr3 = {10, 15, 25, 30, 35};
// using lower_bound() to check if 20 exists
// single occurrence
// prints 2
cout << "The position of 20 using lower_bound "
" (in single occurrence case) : ";
cout << lower_bound(arr1.begin(), arr1.end(), 20)
- arr1.begin();
cout << endl;
// using lower_bound() to check if 20 exists
// multiple occurrence
// prints 2
cout << "The position of 20 using lower_bound "
"(in multiple occurrence case) : ";
cout << lower_bound(arr2.begin(), arr2.end(), 20)
- arr2.begin();
cout << endl;
// using lower_bound() to check if 20 exists
// no occurrence
// prints 2 ( index of next higher)
cout << "The position of 20 using lower_bound "
"(in no occurrence case) : ";
cout << lower_bound(arr3.begin(), arr3.end(), 20)
- arr3.begin();
cout << endl;
}
输出：
The position of 20 using lower_bound (in single occurrence case) : 2
The position of 20 using lower_bound (in multiple occurrence case) : 2
The position of 20 using lower_bound (in no occurrence case) : 2
3. upper_bound(start_ptr，end_ptr，num) ：如果容器包含1次num，则返回指向“下一个比num高的数字的位置”的指针。如果容器包含num的多次出现，则返回指向“比上次出现的num高一个下一个数字的第一个位置”的指针。如果容器不包含num，则返回指向“下一个比num高的数字的位置”的指针。将指针减去到第一位置即“ vect.begin() ”将返回实际索引。
// C++ code to demonstrate the working of upper_bound()
#include
using namespace std;
int main()
{
// initializing vector of integers
// for single occurrence
vector arr1 = {10, 15, 20, 25, 30, 35};
// initializing vector of integers
// for multiple occurrences
vector arr2 = {10, 15, 20, 20, 25, 30, 35};
// initializing vector of integers
// for no occurrence
vector arr3 = {10, 15, 25, 30, 35};
// using upper_bound() to check if 20 exists
// single occurrence
// prints 3
cout << "The position of 20 using upper_bound"
" (in single occurrence case) : ";
cout << upper_bound(arr1.begin(), arr1.end(), 20)
- arr1.begin();
cout << endl;
// using upper_bound() to check if 20 exists
// multiple occurrence
// prints 4
cout << "The position of 20 using upper_bound "
"(in multiple occurrence case) : ";
cout << upper_bound(arr2.begin(), arr2.end(), 20)
- arr2.begin();
cout << endl;
// using upper_bound() to check if 20 exists
// no occurrence
// prints 2 ( index of next higher)
cout << "The position of 20 using upper_bound"
" (in no occurrence case) : ";
cout << upper_bound(arr3.begin(), arr3.end(), 20)
- arr3.begin();
cout << endl;
}
输出：
The position of 20 using upper_bound (in single occurrence case) : 3
The position of 20 using upper_bound (in multiple occurrence case) : 4
The position of 20 using upper_bound (in no occurrence case) : 2
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