Factorial Trailing Zeroes

 

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

Note: Your solution should be in logarithmic time complexity.

只有2 * 5才会产生0只要计算出因子中2和5的个数取小的的那个就好了

public class Solution {
    public int trailingZeroes(int n) {
        int numsOf2 = 0;
        int numsOf5 = 0;
        for(int i = 2; i <= n; i++){
            numsOf2 += get2nums(i);
            numsOf5 += get5nums(i);
        }
        return numsOf2 < numsOf5 ? numsOf2 : numsOf5;
    }
    
    /**
     * 计算num中2作为乘积因子的个数
     * @param num
     * @return
     */
    private int get2nums(int num){
        int numsOf2 = 0;
        if(num % 2 != 0)
            return 0;
        else{
            while(num % 2 == 0){
                num = num / 2;
                numsOf2 ++;
            }
        }
        return numsOf2;
    }
    
    /**
     * 获取5的个数
     * @param num
     * @return
     */
    private int get5nums(int num){
        int numsOf5 = 0;
        if(num % 5 != 0)
            return 0;
        else{
            while(num % 5 == 0){
                num = num / 5;
                numsOf5 ++;
            }
        }
        return numsOf5;
    }
}