一种二分查找变形

原始的二分查找非常简单，只要找到值返回即可；但是如果某值在数组中是重复的，比如 1 3 5 5 5 5 10 11 20这样，要求找到第一个5和最后一个5，就需要对原始的二分查找进行一点变形。

代码如下：

非递归版

int GetFirst_NonRecurse(int a[], int N, int k, int start, int end)
{

	int mid;
	while (start <= end)
	{
		mid = start + (end-start)/2;
		
		if (a[mid] == k)
		{
			if (mid>0&&a[mid-1]!=k || mid==0)
				return mid;
			else
				end = mid - 1;
		}
		else if (a[mid] > k)
			end = mid - 1;
		else
			start = mid + 1;
	}
	return  -1;
}

int GetLast_NonRecurse(int a[], int N, int k, int start, int end)
{
	int mid;
	while (start <= end)
	{
		mid = start + (end - start)/2;
		if (a[mid] == k)
		{ 
			if (mid < N-1 && a[mid+1] != k || mid == N-1)
				return mid;
			else
				start = mid + 1;
		}
		else if (a[mid] > k)
		{
			start = mid + 1 ;
		}
		else
		{
			end = mid - 1;
		}
	}
	return -1;
}

递归版

int GetFirst(int a[], int N, int k, int start, int end)
{
	if (start > end)
		return -1;

	int mid = start + (end-start)/2;

	if (a[mid] == k)
	{
		if (mid>0&&a[mid-1]!=k || mid==0)
			return mid;
		else
			end = mid - 1;
	}
	else if (a[mid] > k)
		end = mid - 1;
	else
		start = mid + 1;

	return GetFirst(a,N,k,start,end);

}
int GetLast(int a[], int N, int k, int start, int end)
{
	if (start > end)
		return -1;

	int mid = start + (end - start)/2;

	if (a[mid] == k)
	{
		if (mid<N-1&&a[mid+1]!=k || mid==N-1)
			return mid;
		else
			start = mid + 1;
	}
	else if (a[mid] > k)
	{
		end = mid - 1;
	}
	else
	{
		start = mid + 1;
	}
	return GetLast(a,N,k,start,end);
}