Java for LeetCode 172 Factorial Trailing Zeroes

標籤：

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

Note: Your solution should be in logarithmic time complexity.

解題思路：

計算n能達到的5的最大次冪，算出在這種情況下能提供的5的個數，然後減去之後遞迴即可，JAVA實現如下：

static public int trailingZeroes(int n) {if(n<25)return n/5;long five=5;int count=0;while(n>=five){five*=5;count++;}int temp=(int) (n/Math.pow(5, count));return countSum(count)*temp+trailingZeroes(n-temp*(int)Math.pow(5, count));}static public int countSum(int count){if(count==1)return 1;else return countSum(count-1)*5+1;}

 

Java for LeetCode 172 Factorial Trailing Zeroes