斐波那契数列算法 Fibonacci Search 优化二分法查找算法

二分法已经非常快速了，但是还是可以改进二分法查找：

不要使用中间元素， 而是使用更为精确的区间，找到关键字在上面区间内，使其收敛更快。

使用Fibonacci序列数字来确定这个区间，就是Fibonacci Search。

二分法请参考：

http://blog.csdn.net/kenden23/article/details/16113241

下面是Wiki对Fibonacci Search的叙述：

To test whether an item is in the list of ordered numbers, follow these steps:

1. Set k = m.
2. If k = 0, stop. There is no match; the item is not in the array.
3. Compare the item against element in F_k−1.
4. If the item matches, stop.
5. If the item is less than entry F_k−1, discard the elements from positions F_k−1+1 to n. Set k=k−1 and return to step 2.
6. If the item is greater than entry F_k−1, discard the elements from positions 1 to F_k−1. Renumber the remaining elements from 1 to F_k−2, set k=k−2, and return to step2.

参考：Wiki

下面就根据这个算法一步一步实现这个程序：

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

void generateFibonacci(vector<int> &fibvec, int n)
{
	if(n>47) n = 47;	//整形数据只能存在47个fibonacci元素了。这里的fibonacci第一个数是0.
	fibvec.push_back(0);
	fibvec.push_back(1);
	for (int i = 2; i < n; i++)
	{
		fibvec.push_back(fibvec[i-1] + fibvec[i-2]);
	}
}

template<typename T>
int fibonacciSearch(vector<T> &vt, T key)
{
	vector<int> fibvec;
	generateFibonacci(fibvec, 47);
	int n = vt.size();

	//1.Set k = m.
	int k = lower_bound(fibvec.begin(), fibvec.end(), n) - fibvec.begin();
	
	//offset is needed for recording the begin point in vt.
	int offset = 0;
	int index = 0;
	//2.If k = 0, stop. There is no match; the item is not in the array.
	while (k>0)
	{
		index = fibvec[--k]+offset;
		//3.Compare the item against element in Fk−1.
		//5.If the item is less than entry Fk−1, discard the elements from positions Fk−1 + 1 to n. Set k = k − 1 and return to step 2.
		//注意：这里的index有可能大于数组最大值n的。因为fibvec[k]有可能远远大于n，那么fibvec[k-1]+offset就有可能也大于n。注意这个特殊情况。比如数组元素有22个，那么就要选取fibvec[k] == 34。那么fibvec[k-1] == 21，第二趟循环的时候offset有可能变成21，那么f[k]==13,f[k-1]==8那么就很容易超标！
		if(index>=n || key < vt[index]) continue;
		//6.If the item is greater than entry Fk−1, discard the elements from positions 1 to Fk−1. Renumber the remaining elements from 1 to Fk−2, set k = k − 2, and return to step 2.
		else if (key > vt[index])
		{
			offset = index;
			k--;
		}
		//4.If the item matches, stop.
		else return index;
		
	}
	return -1;
}

int main()
{
	std::vector<int> vec;

	// set some initial content:
	for (int i=1;i<10;i++) vec.push_back(i<<2);

	vec.resize(7);
	vec.resize(12,80);

	std::cout << "vec contains:";
	for (int i=0;i<vec.size();i++)
		std::cout << ' ' << vec[i];
	std::cout << '\n';
	cout<<endl<<"Fibonacci List:\n";
	vector<int> fibvec;
	generateFibonacci(fibvec, 120);
	for (auto x:fibvec)
		cout<<x<<" ";
	cout<<endl<<endl;;

	cout<<"Fibonacci Search: ";
	int fibnd = fibonacciSearch(vec, 28);
	cout<<fibnd;
	if(fibnd != -1)
		cout<<"\tValue: "<<vec[fibnd]<<endl;
	else cout<<"\tNo Found!"<<endl;

	system("pause");
	return 0;
}

主要注意这个容易出错的地方：这里的index有可能大于数组最大值n的。因为fibvec[k]有可能远远大于n，那么fibvec[k-1]+offset就有可能也大于n。注意这个特殊情况。比如数组元素有22个，那么就要选取fibvec[k] == 34。那么fibvec[k-1] == 21，第二趟循环的时候offset有可能变成21，那么f[k]==13,f[k-1]==8那么就很容易超标！

运行结果：