二元搜索的抽象

什么是二进制搜索算法?
二进制搜索算法用于查找x的某个值，为此需要将某个定义的函数f(x)最大化或最小化。通过将搜索间隔重复分成两半，它经常用于按排序顺序搜索元素。从一个覆盖整个序列的间隔开始，如果搜索关键字的值小于间隔中间的项目，则在间隔的左半部分进行搜索，否则在右半部分进行搜索。重复检查，直到找到该值或间隔为空。
执行二进制搜索的主要条件是序列必须是单调的，即，它必须是递增或递减的。

Monotonic function 
A function f(x) is said to be monotonic if and only if for any x if f(x) returns true, then for any value of y (where y > x) should also return true and similarly if for a certain value of x for which f(x) is false, then for any value z (z < x) the function should also return false. 
 

如何使用Binary Search解决问题：
如果函数是

$f(x) = x^{2}$

而任务是找到x的最大值，以使f(x)小于或等于目标值。我们将搜索给定目标值的时间间隔

$f(x)$

从0到目标值。
然后我们可以使用二进制搜索来解决这个问题，因为该函数

$f(x) = x^{2}$

是单调增加的函数。

Target = 17 
f(x) = x², since the function is monotonic binary search can be applied to it. 
Range of search interval will be [0, target]
Step 1: 
low = 0, high = 17, calculate mid = (low + high)/2 = 8 
Calculate f(8) = 64 which is more than target, so it will return false and high will be updated as high = mid – 1 = 7.
Steps 2: 
low = 0, high = 7, calculate mid = (low + high)/2 = 3 
Calculate f(3) = 9 which is less than target, so it will return true and low will be updated as low = mid + 1 = 4.
Step 3: 
low = 4, high = 7, calculate mid = (low + high)/2 = 5 
Calculate f(5) = 25 which is more than target, so it will return false and high will be updated as high = mid – 1 = 4.
Step 4: 
Now since the range [low, high] converges to a single point i.e 4 so the final result is found, since f(4) = 16 which is the maximum value of the given function less than target. 
 

下面是上述示例的实现：

C++

// C++ program for the above example
  
#include "bits/stdc++.h"
using namespace std;
  
// Function to find X such that it is
// less than the target value and function
// is f(x) = x^2
void findX(int targetValue)
{
  
    // Initialise start and end
    int start = 0, end = targetValue;
    int mid, result;
  
    // Loop till start <= end
    while (start <= end) {
  
        // Find the mid
        mid = start + (end - start) / 2;
  
        // Check for the left half
        if (mid * mid <= targetValue) {
  
            // Store the result
            result = mid;
  
            // Reinitialize the start point
            start = mid + 1;
        }
  
        // Check for the right half
        else {
            end = mid - 1;
        }
    }
  
    // Print the maximum value of x
    // such that x^2 is less than the
    // targetValue
    cout << result << endl;
}
  
// Driver Code
int main()
{
    // Given targetValue;
    int targetValue = 81;
  
    // Function Call
    findX(targetValue);
}

Java

// Java program for
// the above example
import java.util.*;
class GFG{
  
// Function to find X such
// that it is less than the
// target value and function
// is f(x) = x^2
static void findX(int targetValue)
{
      
    // Initialise start and end
    int start = 0, end = targetValue;
    int mid = 0, result = 0;
  
    // Loop till start <= end
    while (start <= end) 
    {
          
        // Find the mid
        mid = start + (end - start) / 2;
  
        // Check for the left half
        if (mid * mid <= targetValue) 
        {
  
            // Store the result
            result = mid;
  
            // Reinitialize the start point
            start = mid + 1;
        }
  
        // Check for the right half
        else
        {
            end = mid - 1;
        }
    }
  
    // Print the maximum value of x
    // such that x^2 is less than the
    // targetValue
    System.out.print(result + "\n");
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given targetValue;
    int targetValue = 81;
  
    // Function call
    findX(targetValue);
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program for
# the above example
  
# Function to find X such
# that it is less than the
# target value and function
# is f(x) = x ^ 2
def findX(targetValue):
    
    # Initialise start and end
    start = 0;
    end = targetValue;
    mid = 0;
  
    # Loop till start <= end
    while (start <= end):
        
        # Find the mid
        mid = start + (end - start) // 2;
  
        # Check for the left half
        if (mid * mid <= targetValue):
            result = mid
            start = mid + 1;
  
        # Check for the right half
        else:
            end = mid - 1;
  
    # Print maximum value of x
    # such that x ^ 2 is less than the
    # targetValue
    print(result);
  
# Driver Code
if __name__ == '__main__':
    
    # Given targetValue;
    targetValue = 81;
  
    # Function Call
    findX(targetValue);
  
# This code is contributed by Rajput-Ji

C#

// C# program for
// the above example
using System;
class GFG{
  
// Function to find X such
// that it is less than the
// target value and function
// is f(x) = x^2
static void findX(int targetValue)
{
      
    // Initialise start and end
    int start = 0, end = targetValue;
    int mid = 0, result = 0;
  
    // Loop till start <= end
    while (start <= end)
    {
          
        // Find the mid
        mid = start + (end - start) / 2;
  
        // Check for the left half
        if (mid * mid <= targetValue)
        {
  
            // Store the result
            result = mid;
  
            // Reinitialize the start point
            start = mid + 1;
        }
  
        // Check for the right half
        else
        {
            end = mid - 1;
        }
    }
      
    // Print the maximum value of x
    // such that x^2 is less than the
    // targetValue
    Console.Write(result + "\n");
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given targetValue;
    int targetValue = 81;
  
    // Function Call
    findX(targetValue);
}
}
  
// This code is contributed by 29AjayKumar

输出：

9

上述二进制搜索算法最多需要O(log N)比较才能找到小于或等于目标值的最大值。函数f(x)= x ^2的值不需要多次求值。
时间复杂度： O(logN)
辅助空间： O(1)