典型递归算法例子-------Python实现

#!/usr/bin/python
#coding=utf-8
'''
Created on 2012-9-25

@author: linzuxin
'''

status = {}

def factorial(n):
    '''
            输入一个数字，求其阶乘。status是一个hash,用来存储中间结果，避免重复计算
    @param n: 要求的数字
    @return: n的阶乘 
    '''
    if n == 0 or n == 1:
        status[0] = 1
        status[1] = 1
        return 1
    elif status.has_key(n):
        return status[n]
    else:
        status[n] = factorial(n - 1) * n
        return status[n] 
    
print factorial(100)

#!/usr/bin/python
#coding=utf-8
'''
Created on 2012-9-25

@author: linzuxin
'''
status = {}

def fibonacci(n):
    '''
            输入一个数字，求其斐波那契数。status是一个hash,用来存储中间结果，避免重复计算
    @param n: 要求的数字
    @return: n的斐波那契数 
    '''
    if n == 0 or n == 1:
        status[0] = 1
        status[1] = 1
        return 1
    elif status.has_key(n):
        return status[n]
    else:
        status[n] = fibonacci(n - 2) + fibonacci(n - 1)
        return status[n]

print fibonacci(100)

#!/usr/bin/python
#coding=utf-8
'''
Created on 2012-9-26

@author: linzuxin
'''
def permutation(array, start, end):
    '''
            输入一个数组，然后输出其全排列
    @param array: 要求全排列的数组
    @param start: 数组的开始下标 
    @param end: 数组的结束下标
    '''
    if start == end:
        for i in array:
            print i,
            
        print
    else:
        for i in range(start, end + 1):
            array[start], array[i] = array[i], array[start]
            permutation(array, start + 1, end)
            array[start], array[i] = array[i], array[start]

array = [1, 2, 3]            
permutation(array, 0, len(array) - 1)

#!/usr/bin/python
#coding=utf-8
'''
Created on 2012-9-26

@author: linzuxin
'''
status = {}

def divideInt(n, m):
    '''
            输入一个数字，求其整数划分种类
    @param n: 要求的数字
    @param m: 划分中最大的加数不大于m
    @return: 划分总数
    '''
    try:
        return status[(n, m)]
    except KeyError:
        if n == 1 or m == 1:
            status[(n, m)] = 1
            return status[(n, m)]
        elif m > n:
            status[(n, m)] = divideInt(n, n)
            return status[(n, m)]
        elif m == n:
            status[(n, m)] = 1 + divideInt(n, n - 1)
            return status[(n, m)]
        else:
            status[(n, m)] = divideInt(n - m, m) + divideInt(n, m - 1)
            return status[(n, m)]
    
print divideInt(500, 500)

#!/usr/bin/python
#coding=utf-8
'''
Created on 2012-9-26

@author: linzuxin
'''

count = 0
def hanoi(n, start, end, middle):
    '''
            输入汉诺塔个数，求出其需要移动的次数
    @param start: 开始汉诺塔的位置
    @param end: 最终汉诺塔的位置
    @param middle: 用来过渡的位置
    '''
    global count
    if n == 1:
        print start, ' => ', end
        count += 1
    else:
        hanoi(n - 1, start, middle, end)
        print start, ' => ', end
        count += 1
        hanoi(n - 1, middle, end, start) 
        
hanoi(2, 'A', 'B', 'C')
print count