【leetcode】Factorial Trailing Zeroes
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Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
思路:
阿里算法工程师笔试的一个选择题。其实就是计算其中5的个数就行。算算从1到n有几个5的倍数。但是这个方法会超时(方法二),方法一思路是一样的,不会超时,因为n一直减少。
方法1:
class Solution {public: int trailingZeroes(int n) { if(n==0) return 0; int res=0; while(n) { res=res+n/5; n=n/5; } return res; }};
方法2:
class Solution {public: int trailingZeroes(int n) { if(n==0) return 0; int res=0; for(int i=1;i<=n;i++) { int temp=i; while(temp%5==0) { res++; temp=temp/5; } } return res; }};
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