学步园

编程之美3.18后面的二分查找题解：

#include<stdlib.h>
#include<stdio.h>
//最大索引，等于k
int bisearch(int* arr, int b, int e, int v)
{
	int minIndex = b, maxIndex = e, midIndex = 0;
	while (minIndex < maxIndex - 1)
	{
		midIndex = minIndex + ((maxIndex - minIndex) >> 1);
		if (arr[midIndex] <= v) minIndex = midIndex;
		else maxIndex = midIndex - 1;
	}
	if (arr[maxIndex] == v) return maxIndex;
	else if (arr[minIndex] == v) return minIndex;
	else return -1; 
}
//任意一个等于V的索引
int bisearch2(int* arr, int b, int e, int v)
{
	int minIndex = b, maxIndex = e, midIndex = 0;
	while (minIndex <= maxIndex)
	{
		midIndex = minIndex + ((maxIndex - minIndex) >> 1);
		if (arr[midIndex] == v) return midIndex;
		else if (arr[midIndex] < v) minIndex = midIndex + 1;
		else maxIndex = midIndex - 1;
	}
	return -1;
}
//最小索引，等于k
int bisearch3(int* arr, int b, int e, int v)
{
	int minIndex = b, maxIndex = e, midIndex = 0;
	while (minIndex < maxIndex - 1)
	{
		midIndex = minIndex + ((maxIndex - minIndex) >> 1);
		if (arr[midIndex] >= v) maxIndex = midIndex;
		else minIndex = midIndex + 1;
	}
	if (arr[minIndex] == v) return minIndex;
	else if (arr[maxIndex] == v) return maxIndex;
	return -1;
}
//最小的索引，大于k
int bisearch4(int* arr, int b, int e, int v)
{
	int minIndex = b, maxIndex = e, midIndex = 0;
	while (minIndex <= maxIndex - 1)
	{
		midIndex = minIndex + ((maxIndex - minIndex) >> 1);
		if (arr[midIndex] > v) maxIndex = midIndex;
		else minIndex = midIndex + 1;
	}
	if (arr[maxIndex] > v) return maxIndex;
	return -1;
}
//最大的索引，小于k
int bisearch5(int* arr, int b, int e, int v)
{
	int minIndex = b, maxIndex = e, midIndex = 0;
	while (minIndex < maxIndex - 1)
	{
		midIndex = minIndex + ((maxIndex - minIndex) >> 1);
		if (arr[midIndex] >= v) maxIndex = midIndex - 1;
		else minIndex = midIndex;
	}
	if (arr[maxIndex] < v) return maxIndex;
	if (arr[minIndex] < v) return minIndex;
	return -1;
}
int main()
{
	int arr[] = {1,2,3,4,5,5,5,6,6,7,8,9,9};
	int k;
	while(1)
	{
		scanf("%d",&k);
		printf("%d,%d\n",sizeof(arr)/sizeof(arr[0]),bisearch5(arr, 0, sizeof(arr)/sizeof(arr[0]) - 1,k));
	}
	return 0;
}