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用二分法实现一个有序数组的查找

1.直接写代码

#include <stdio.h>

int main()

{
int arr[] = { 1,2,3,4,5,6,7,8,9,10 };
int k = 7;
int sz = sizeof(arr) / sizeof(arr[0]);
int ret = binary_search(arr, k, sz);
if (ret == -1)
{
printf("找不到\n");
}
else
{
printf("找到了,下标是:%d\n", ret);
}
return 0;
}
int main()
{
int arr [] = { 1,2,3,4,5,6,7,8,9,10 };
int k = 7;
int left = 0;
int sz = sizeof(arr) / sizeof(arr[0]);
int right =sz-1;

while (left <= right)
{
int mid = (left + right) / 2;
if (arr[mid] > k)
{
right = mid - 1;
}
else if (arr[mid] < k)
{
left = mid + 1;
}
else
{
printf("找到了,下标是:%d\n", mid);
break;
}
}
if (left > right)
{
printf("找不到\n");
}
return 0;

用二分法实现一个有序数组的查找_#include

2.定义一个函数实现二分法查找

#include <stdio.h>

int binary_search(int arr[], int k, int sz)

{
int left = 0;
int right = sz - 1;

while (left<=right)
{
int mid = (left + right) / 2;
if (arr[mid] > k)
{
right = mid - 1;
}
else if (arr[mid] < k)
{
left = mid + 1;
}
else
{
return mid;
}
}
if (left > right)
{
return - 1;
}
}​

int main()

{

int arr[] = { 1,2,3,4,5,6,7,8,9,10 };

int k = 7;

int sz = sizeof(arr) / sizeof(arr[0]);

int ret = binary_search(arr, k, sz);

if (ret == -1)

{

 printf("找不到\n");

}

else

{

 printf("找到了,下标是:%d\n", ret);

}

return 0;

}

用二分法实现一个有序数组的查找_#include_02

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