维基上的代码：
int binary_search(int A[], int key, int imin, int imax) { // continue searching while [imin,imax] is not empty while (imax >= imin) { // calculate the midpoint for roughly equal partition int imid = midpoint(imin, imax); if(A[imid] == key) // key found at index imid return imid; // determine which subarray to search else if (A[imid] < key) // change min index to search upper subarray imin = imid + 1; else // change max index to search lower subarray imax = imid - 1; } // key was not found return KEY_NOT_FOUND; }下面使用了两个算法。
int binary_search()
{
int size ;
char keys[] = "9290";
char words[][6] ={"1","2","21","22","222","3","5","66","90","900","91","929"};
size = sizeof(words) / sizeof(words[0]);
printf("第一个算法 sizeof = %d\n", size);
int head = 0;
int tail = size;
int cursor = (head + tail) / 2;
int result = 999;
int count = 0;
printf("keys= \"%s\"\n", keys);
printf("words[i][] = \n");
for (int i = 0; i < size; ++i) printf("[%d]=\"%s\" ", i, words[i]);
do
{
result = strcmp(keys, words[cursor]);
printf("\ncount = %d cursor = %d head = %d , tail = %d , \n", count++ ,cursor, head, tail);
printf("words[%d] = \"%s\" , result = %d \n\n ", cursor, words[cursor] ,result);
if(result > 0)
{
head = cursor;
}
else if (result < 0)
{
tail = cursor;
}
else
{
break;//找到匹配的key按键组合
}
cursor = (head + tail) / 2;
}
while((head < cursor) && (cursor <= tail)); //((head <= cursor) && (cursor < tail))
printf("\n------------------\ncursor = %d head = %d , tail = %d , \n", cursor, head, tail);
printf("words[%d] = \"%s\" , result = %d \n\n", cursor, words[cursor], result);
printf("====================\n");
int left = 0, right = size - 1, middle = 0; //, result = 0;
count = 0;
printf("第二个算法 sizeof = %d\n", size);
printf("keys= \"%s\"\n", keys);
printf("words[i][] = \n");
for (int i = 0; i < size; ++i) printf("[%d]=\"%s\" ", i, words[i]);
printf("\nmiddle = %d, left = %d , right = %d , \n", middle, left, right);
while (left <= right)
{
middle = (left + right) >> 1;
result = strcmp(keys, words[middle]);
printf("count = %d middle = %d, left = %d , right = %d , \n", count++, middle , left, right);
printf("words[%d] = \"%s\" , result = %d \n\n", middle , words[middle], result);
if (result > 0)
{
left = middle + 1;
}
else if (result < 0) right = middle - 1;
else break;
}
printf("---------------------\nmiddle = %d, left = %d , right = %d , \n", middle, left, right);
printf("words[%d] = \"%s\" , result = %d \n ", middle , words[middle], result);
return cursor;
}
第一个算法 sizeof = 12
keys= "9290"
words[i][] =
[0]="1" [1]="2" [2]="21" [3]="22" [4]="222" [5]="3" [6]="5" [7]="66" [8]="90" [9
]="900" [10]="91" [11]="929"
count = 0 cursor = 6 head = 0 , tail = 12 ,
words[6] = "5" , result = 1
count = 1 cursor = 9 head = 6 , tail = 12 ,
words[9] = "900" , result = 1
count = 2 cursor = 10 head = 9 , tail = 12 ,
words[10] = "91" , result = 1
count = 3 cursor = 11 head = 10 , tail = 12 ,
words[11] = "929" , result = 1
------------------
cursor = 11 head = 11 , tail = 12 ,
words[11] = "929" , result = 1
====================
第二个算法 sizeof = 12
keys= "9290"
words[i][] =
[0]="1" [1]="2" [2]="21" [3]="22" [4]="222" [5]="3" [6]="5" [7]="66" [8]="90" [9
]="900" [10]="91" [11]="929"
middle = 0, left = 0 , right = 12 ,
count = 0 middle = 6, left = 0 , right = 12 ,
words[6] = "5" , result = 1
count = 1 middle = 9, left = 7 , right = 12 ,
words[9] = "900" , result = 1
count = 2 middle = 11, left = 10 , right = 12 ,
words[11] = "929" , result = 1
---------------------
middle = 12, left = 12 , right = 12 ,
words[12] = "烫烫烫烫9290" , result = 1
请按任意键继续. . .
忙于工作，有机会再研究效率的问题。