C语言实现:
#include <stdio.h>
int binsearch(int x, int v[], int n)
{
int low, high, mid;
low = 0;
high = n - 1;
while (low <= high){
mid = (low + high) / 2;
if (x < v[mid])
high = mid - 1;
else if (x > v[mid])
low = mid + 1;
else
return mid;
}
return -1;
}
main()
{
int arr[] = {1,2, 5, 6, 8, 9, 10, 23, 56};
printf("%d\n", binsearch(8, arr, 9));
}
int binsearch(int x, int v[], int n)
{
int low, high, mid;
low = 0;
high = n - 1;
while (low <= high){
mid = (low + high) / 2;
if (x < v[mid])
high = mid - 1;
else if (x > v[mid])
low = mid + 1;
else
return mid;
}
return -1;
}
main()
{
int arr[] = {1,2, 5, 6, 8, 9, 10, 23, 56};
printf("%d\n", binsearch(8, arr, 9));
}
二分查找的基本思想是将n个元素分成大致相等的两部分,去a[n/2]与x做比较,如果x=a[n/2],则找到x,算法中止;如果x<a[n/2],则只要在数组a的左半部分继续搜索x,如果x>a[n/2],则只要在数组a的右半部搜索x.
时间复杂度无非就是while循环的次数!
总共有n个元素,
渐渐跟下去就是n,n/2,n/4,....n/2^k,其中k就是循环的次数
由于你n/2^k取整后>=1
即令n/2^k=1
可得k=log2n,(是以2为底,n的对数)
所以时间复杂度可以表示O()=O(logn)