20170311-leetcode-Fractial Trailing Zeroes
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1.Fractorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
https://leetcode.com/problems/factorial-trailing-zeroes/?tab=Description
解读
求一个整数的阶乘尾部零的个数,夹杂在中间的不算,并且要求算法复杂度为O(log(N))
2.Solution
拿到这个题的时候,如果没有什么思路,可以先把阶乘打印出来找规律
1!=1 zeros:02!=2 zeros:03!=6 zeros:04!=24 zeros:05!=120 zeros:16!=720 zeros:17!=5040 zeros:18!=40320 zeros:19!=362880 zeros:110!=3628800 zeros:211!=39916800 zeros:212!=479001600 zeros:213!=6227020800 zeros:214!=87178291200 zeros:215!=1307674368000 zeros:316!=20922789888000 zeros:317!=355687428096000 zeros:318!=6402373705728000 zeros:319!=121645100408832000 zeros:320!=2432902008176640000 zeros:421!=51090942171709440000 zeros:422!=1124000727777607680000 zeros:423!=25852016738884976640000 zeros:424!=620448401733239439360000 zeros:425!=15511210043330985984000000 zeros:6
可以发现
前面24个数字都是n/5取整的结果,25增加了1,可以尝试输出一下50,75,100,125等数字,都会增加0的个数。
如果要出现0,则必须出现5,5与偶数结合就出现了0,5的个数决定了0的个数,而25,50,75,可以拆分成5X5,5X5X2,5X5X5,因此0的个数在原来的基础上在增加
class Solution(object): def trailingZeroes(self, n): cnt=0 while n>4: cnt+=int(n/5) n=int(n/5) return cnt
0 0
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