本程序可根据给定的字符字典,穷举指定长度的所有字符串: def get_pwd(str, num): if(num == 1): for x in str: yield x else: for x in str: for y in get_pwd(str, num-1): yield x+y strKey="abc"for x in get_pwd(strKey,3):
本程序可根据给定的字符字典,穷举指定长度的所有字符串:
def get_pwd(str, num): if(num == 1): for x in str: yield x else: for x in str: for y in get_pwd(str, num-1): yield x+y strKey="abc" for x in get_pwd(strKey,3): print x
结果:
aaa aab aac aba abb abc aca acb acc baa bab bac bba bbb bbc bca bcb bcc caa cab cac cba cbb cbc cca ccb ccc
本程序占用内存小,生成速度快,欢迎尝试!!!
补充知识:Python 穷举法, 二分法 与牛顿-拉夫逊方法求解平方根的性能对比
穷举法, 二分法 与牛顿-拉夫逊方法求解平方根的优劣,从左到右依次递优。
经过测试,穷举法基本超过 1 分钟,还没有出数据;
二分法只要区区1秒不到就出结果了。
牛顿-拉夫逊是秒出,没有任何的停顿。
numberTarget =int(input("Please enter a number:")) numberSqureRoot = 0 while(numberSqureRoot<abs(numberTarget)): if numberSqureRoot**2 >= abs(numberTarget): break numberSqureRoot = numberSqureRoot + 1 if numberSqureRoot**2 != numberTarget: print("Your number %s is not a perfect squre, the square root is %s " % ( numberTarget,numberSqureRoot) ) else: if numberTarget < 0 : numberSqureRoot = -numberSqureRoot print("Your number %s is a perfect squre, the square root is %s " % ( numberTarget, numberSqureRoot)) print("now we begin to calculate the binary search...") numberTarget=int(input("Please enter the number for binary search...")) numberSqureRoot = 0 lowValue = 0.0 highValue=numberTarget*1.0 epsilon = 0.01 numberSqureRoot = (highValue + lowValue)/2 while abs(numberSqureRoot**2 - numberTarget) >=epsilon: print("lowValue:%s, highValue:%s, currentValue:%s"%(lowValue,highValue,numberSqureRoot)) if numberSqureRoot**2<numberTarget: lowValue=numberSqureRoot else: highValue=numberSqureRoot numberSqureRoot = (lowValue+highValue) /2 print("The number %s has the squre root as %s " %(numberTarget,numberSqureRoot)) print("now we begin to calculate the newTon search...") numberTarget=int(input("Please enter the number for newTon search...")) numberSqureRoot = 0 epsilon = 0.01 k=numberTarget numberSqureRoot = k/2.0 while( abs(numberSqureRoot*numberSqureRoot - k)>=epsilon): numberSqureRoot=numberSqureRoot-(((numberSqureRoot**2) - k)/(2*numberSqureRoot)) print("squre root of %s is %s " %(numberTarget,numberSqureRoot))
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