172. 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：有多少个0就是有多少个5. 零的产生就是5和2或者4的乘积，15—》 15*12/14   10*12/14 5*2/4 -》3个。 25拆成5*5所以会多一个。

public class Solution {
    public int trailingZeroes(int n) {
        if(n<=0)return 0;
        int res=0;
        while(n>0){
            res+=n/5;
            n/=5;
        }
        return res;
    }
}

 

转载于:https://www.cnblogs.com/Machelsky/p/5955468.html