C语言二分查找法（指针和数组实现）

/*
* 编写一个函数，对一个已排序的整数表执行二分查找。
* 函数的输入包括各异指向表头的指针，表中的元素个数，以及待查找的数值。
* 函数的输出时一个指向满足查找要求的元素的指针，当未查找到要求的数值时，输出一个NULL指针
* 用两个版本实现，一个用的是数组小标，第二个用的是指针
* 他们均采用了不对称边界
*
Copyright (c) 2012 LiMing
Author:        LiMing
2012-06-21
referenced C Traps and Pitfaills Chinese Edition
Page 132-137
*
* 查找的元素为x，数组下表是k，开始时0 <= k < n
* 接下来缩小范围lo <= k < hi,
* if lo equals hi, we can justify the element "x" is not in the array

* */
#include <stdio.h>

int array[] =
{
0,1,2,3,4,5,6,7
};

int *bsearch_01(int *t, int n, int x);

int *bsearch_01(int *t, int n, int x)
{
int lo = 0;
int hi = n;

while(lo < hi)
{
//int mid = (hi + lo) / 2;
int mid = (hi + lo) >> 1;

if(x < t[mid])
hi = mid;
else if(x > t[mid])
lo = mid + 1;
else
return t + mid;
}
return NULL;
}

int *bsearch_02(int *t, int n, int x);

int *bsearch_02(int *t, int n, int x)
{
int lo = 0;
int hi = n;

while(lo < hi)
{
//int mid = (hi + lo) / 2;
int mid = (hi + lo) >> 1;
int *p = t + mid;        //用指针变量p存储t+mid的值，这样就不需要每次都重新计算

if(x < *p)
hi = mid;
else if(x > *p)
lo = mid + 1;
else
return p;
}
return NULL;
}

//进一步减少寻址运算
/*
* Suppose we want to reduce address arithmetic still further
* by using pointers instead of subscripts throughout the program.
*
* */
int *bsearch_03(int *t, int n, int x);

int *bsearch_03(int *t, int n, int x)
{
int *lo = t;
int *hi = t + n;

while(lo < hi)
{
//int mid = (hi + lo) / 2;
int *mid = lo + ((hi - lo) >> 1);

if(x < *mid)
hi = mid;
else if(x > *mid)
lo = mid + 1;
else
return mid;
}
return NULL;
}

/*
* The resulting program looks somewhat neater because of the symmetry
* */
int *bsearch_04(int *t, int n, int x);

int *bsearch_04(int *t, int n, int x)
{
int lo = 0;
int hi = n - 1;

while(lo <= hi)
{
//int mid = (hi + lo) / 2;
int mid = (hi + lo) >> 1;

if(x < t[mid])
hi = mid - 1;
else if(x > t[mid])
lo = mid + 1;
else
return t + mid;
}
return NULL;
}

int main(int argc, char **argv)
{
int * ret = NULL;
int *ret2 = NULL;
int *ret3 = NULL;
int *ret4 = NULL;

ret = bsearch_01(array, 8, 3);
ret2 = bsearch_02(array, 8, 6);
ret3 = bsearch_03(array, 8, 4);
ret4 = bsearch_04(array, 8, 2);
printf("The result is %d\n", *ret);
printf("The result is %d\n", *ret2);
printf("The result is %d\n", *ret3);
printf("The result is %d\n", *ret4);

printf("hello world\n");
return 0;
}