每天一道LeetCode--172. Factorial Trailing Zeroes

破玉發表於2016-11-20

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

 此題是求階乘後面零的個數。

public class Solution {
    public int trailingZeroes(int n) {
        int t=0;
        while(n!=0){
            n/=5;
            t+=n;
        }
        return t;
    }
}

 

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