leetcode 172: Factorial Trailing Zeroes

Factorial Trailing Zeroes

Total Accepted: 28 Total Submissions: 69

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

Note: Your solution should be in polynomial time complexity.

[分析]

分解因子, 当且仅当 因子中出现 一对 (2,5)时, 最后结果会增加一个 trailing zero.

1.  2的个数永远多于5个个数.

2.  计算5的个数时, 最简单的方法是 SUM(N/5^1,  N/5^2, N/5^3...)

[注意事项]

[CODE]

public class Solution { 
    public int trailingZeroes(int n) { 
        if(n<1) return 0; 
        int c = 0; 
         
        while(n/5 != 0) {  
            n /= 5; 
            c += n; 
        } 
         
        return c; 
    } 
}