Step 1 : Find the middle element of array. using , middle = initial_value + end_value / 2 ; Step 2 : If middle = element, return ‘element found’ and index. Step 3 : if middle > element, call the function with end_value = middle - 1 . Step 4 : if middle < element, call the function with start_value = middle + 1 . Step 5 : exit.

二分搜索算法的实现函数使用了反复调用函数。这个调用可以有两种类型：

• 迭代
• 递归

迭代调用是多次循环同一段代码。

递归调用是反复调用同一个函数。

使用迭代调用实现二分搜索的程序

示例

演示

#include <stdio.h>
int iterativeBinarySearch(int array[], int start_index, int end_index, int element){
   while (start_index <= end_index){
      int middle = start_index + (end_index- start_index )/2;
      if (array[middle] == element)
         return middle;
      if (array[middle] < element)
         start_index = middle + 1;
      else
         end_index = middle - 1;
   }
   return -1;
}
int main(void){
   int array[] = {1, 4, 7, 9, 16, 56, 70};
   int n = 7;
   int element = 16;
   int found_index = iterativeBinarySearch(array, 0, n-1, element);
   if(found_index == -1 ) {
      printf("Element not found in the array ");
   }
   else {
      printf("Element found at index : %d",found_index);
   }
   return 0;
}

输出

Element found at index : 4

使用递归调用实现二分查找的程序

示例

在线演示

#include <stdio.h>
int recursiveBinarySearch(int array[], int start_index, int end_index, int element){
   if (end_index >= start_index){
      int middle = start_index + (end_index - start_index )/2;
      if (array[middle] == element)
         return middle;
      if (array[middle] > element)
         return recursiveBinarySearch(array, start_index, middle-1, element);
      return recursiveBinarySearch(array, middle+1, end_index, element);
   }
   return -1;
}
int main(void){
   int array[] = {1, 4, 7, 9, 16, 56, 70};
   int n = 7;
   int element = 9;
   int found_index = recursiveBinarySearch(array, 0, n-1, element);
   if(found_index == -1 ) {
      printf("Element not found in the array ");
   }
   else {
      printf("Element found at index : %d",found_index);
   }
   return 0;
}

输出

Element found at index : 3