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该怎么办？
#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;
}
问题是您的while测试。您有：
while(low<high) {
...
}
如果所需值在low == high处，则将失败。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。
编辑：请参阅@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;
}
对于外壳程序的数组元素数目的入门，请使用size_t类型。类型为int的对象可以很小，以容纳数组中元素的数量。
此循环条件
int high=size-1;
while(low<high){
//...
不正确。例如，假设数组只有一个元素。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。
您需要更改条件，例如
while ( !( high < low ) )
//...
else语句中的this if语句
else if(element==arr[mid]){
是多余的。你可以写
else // if(element==arr[mid]){
如果将执行二进制搜索的代码放在单独的函数中会更好。
这里是一个演示程序，显示了如何编写这样的函数。
#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;
}
其输出可能例如以以下方式查看
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