Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
就是返回 n! 这个数里“零的个数”
public class Solution {public int trailingZeroes(int n) {int result = 0;if (n < 0) {return -1;} else if (n < 5) {return 0;} else {while (n != 0) {n = n/5;result += n;}}return result;}
}