Question

我无法解决逻辑错误，因为我不知道这段代码有什么问题。每次输入时，都会显示“找不到元素”。如果有人可以帮助我，我将非常感激。同样在这段代码中，我假设我们将数组的大小作为奇数，如果我们决定将偶数作为大小怎么办？

#include<stdio.h>
int main(){
  int size;
  printf("Enter the number of elemets(odd number) : ");
  scanf("%d",&size);
  int arr[size];
  printf("Enter the elements in ascending order : ");
  for(int i=0;i<size;i++){
    scanf("%d",&arr[i]);
  }
  int element;
  int flag=0;
  printf("Enter element to be found : ");
  scanf("%d",&element);
  int low=0;
  int high=size-1;
  while(low<high){
    int mid=(low+high)/2;

    if(element<arr[mid]){
      high=mid-1;
    }
    else if(element>arr[mid]){
      low=mid+1;
    }
    else if(element==arr[mid]){
      printf("Element %d found at pos %d ",element,mid);
      flag=1;
      break;
    }
  }
  if(flag==0){
    printf("Element not found");
  }

  return 0;
}

Answer 1

问题是您的while测试。你有：

while(low<high) {
    ...
}

如果low == high处的期望值在该位置，则此操作将失败。通过将测试更改为：

while(low <= high) {
    ...
}

这是修复它所需的全部。您无需添加任何特殊情况即可对其进行“修复”。只要确保您的数组按升序排列就可以了。

Answer 2

编辑：请参考@TomKarzes的更好答案

我的旧答案是：

您错过了高==低的边界情况

#include<stdio.h>
int main(){
  int size;
  printf("Enter the number of elements(odd number) : ");
  scanf("%d",&size);
  int arr[size];
  printf("Enter the elements in ascending order : ");
  for(int i=0;i<size;i++){
    scanf("%d",&arr[i]);
  }
  int element;
  int flag=0;
  printf("Enter element to be found : ");
  scanf("%d",&element);
  int low=0;
  int high=size-1;
  while(low<high){
    int mid=(low+high)/2;

    if(element<arr[mid]){
      high=mid-1;
    }
    else if(element>arr[mid]){
      low=mid+1;
    }
    else if(element==arr[mid]){
      printf("Element %d found at pos %d ",element,mid);
      flag=1;
      break;
    }
  }
  if(low==high && arr[low]==element) //Added 1 extra condition check that you missed
  {
    printf("Element %d found at pos %d ",element,low);
    flag=1;
  }
  if(flag==0){
    printf("Element not found");
  }

  return 0;
}

Answer 3

对于壳类数组元素数量的入门级，请使用类型size_t。类型为int的对象可以很小，以容纳数组中元素的数量。

循环的这种情况

int high=size-1;
while(low<high){
//...

不正确。例如，假设该数组只有一个元素。在这种情况下，high等于0，因此由于其初始化而等于left

int high=size-1;

因此，循环不会迭代，尽管数组的第一个和单个元素实际上等于该数字，但您会得到在数组中找不到输入的数字。

您需要像这样更改条件

while ( !( high < low ) )
//...

else语句中的if语句

else if(element==arr[mid]){

是多余的。你可以写

else // if(element==arr[mid]){

最好将执行二进制搜索的代码放在单独的函数中。

这里是一个演示程序，显示了如何编写此函数。

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int binary_search( const int a[], size_t n, int value )
{
    size_t left = 0, right = n;  
    int found = 0;

    while ( !found && left != right )
    {
        size_t middle = left + ( right - left ) / 2;

        if (  value < a[middle] )
        {
            right = middle;
        }
        else if ( a[middle] < value )
        {
            left = middle + 1;
        }
        else
        {
            found = 1;
        }
    }

    return found;
}

int cmp( const void *a, const void *b )
{
    int left  = *( const int * )a;
    int right = *( const int * )b;

    return ( right < left ) - ( left < right );
}

int main(void) 
{
    const size_t N = 15;

    srand( ( unsigned int )time( NULL ) );

    for ( size_t i = 0; i < N; i++ )
    {
        size_t n = rand() % N + 1;

        int a[n];

        for ( size_t j = 0; j < n; j++ ) a[j] = rand() % N;

        qsort( a, n, sizeof( int ), cmp );

        for ( size_t j = 0; j < n; j++ )
        {
            printf( "%d ", a[j] );
        }
        putchar( '\n' );

        int value = rand() % N;

        printf( "The value %d is %sfound in the array\n",
                value, binary_search( a, n, value ) == 1 ? "" : "not " );
    }

    return 0;
}

例如，其输出可能采用以下方式

0 2 2 3 4 5 7 7 8 9 10 12 13 13 
The value 5 is found in the array
4 8 12 
The value 10 is not found in the array
1 2 6 8 8 8 9 9 9 12 12 13 
The value 10 is not found in the array
2 3 5 5 7 7 7 9 10 14 
The value 11 is not found in the array
0 1 1 5 6 10 11 13 13 13 
The value 7 is not found in the array
0 3 3 3 4 8 8 10 11 12 14 14 14 14 
The value 3 is found in the array
0 5 5 10 11 11 12 13 13 14 14 
The value 12 is found in the array
3 4 5 7 10 13 14 14 14 
The value 14 is found in the array
0 3 3 7 
The value 2 is not found in the array
1 6 9 
The value 10 is not found in the array
2 2 3 3 4 4 4 5 5 6 8 8 9 13 13 
The value 11 is not found in the array
11 11 13 
The value 11 is found in the array
0 0 0 1 2 5 5 5 7 7 8 9 12 12 14 
The value 6 is not found in the array
8 8 13 
The value 1 is not found in the array
2 2 4 4 5 9 9 10 12 12 13 13 14 14 
The value 14 is found in the array

C中的二进制搜索代码无法正常工作

3 个答案: