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折半查找
条件: 必须是有序的元素队列
目的:查找该元素队列中是否有该元素,查找成功(找到目标元素)返回元素位置,失败(左右边界出现交叉)返回-1
int binsearch(int list[],int searchnum,int left,int right)
{
// search list[0]<=list[1]...<=list[n-1] for searchnum.
// return its position if found otherwise return -1;
int middle;
while(left <= right){
middle=(left+right)/2;
switch(COMPARE(list[middle],searchnum)){
case -1: left=middle+1; break;
case 0: return middle;
case 1: right=middle-1;
}
}
return -1;
}
递归调用
int binsearch(int list[],int searchnum,int left,int right)
{
/*search list[0]<=list[1]...<=list[n-1] for searchnum.
return its position if found otherwise return -1;*/
printf("\nlrm: %d %d \n",left,right);
printf("\nlrm: %d %d \n",list[left],list[right]);
int middle;
if(left <= right){
middle=(left+right)/2;
printf("\nm ,middle: %d %d \n",middle,list[middle]);
switch(COMPARE(list[middle],searchnum)){
case -1: return binsearch( list, searchnum, middle+1, right) ;
case 0: return middle;
case 1: return binsearch( list, searchnum, left, middle-1) ;
}
}
return -1;
}
折半查找