Given an integer n, return the number of trailing zeroes inn!.
Note: Your solution should be in logarithmic time complexity.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
题目解析:
这道题是问n!的结果后面有几个0的, 可以转化成n!里面有几个5的 (注意25, 125..这种都是5的平方啊立方啊这种要注意一下)
1 public int trailingZeroes(int n) { 2 if(n <= 0) 3 return 0; 4 int num = 0; 5 while(n != 0){ 6 n = n / 5; 7 num += n; 8 } 9 return num; 10 }