leetcode--Factorial Trailing Zeroes
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Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
public class Solution { /** * 本质就是计算因子中5的个数 * n/5可以求出1-n有多少个数能被5整除 * 接着求1-n/5有多少个数能被5整除,也就是能整除25,以此类推 * @param n * @return */public int trailingZeroes(int n) { int counter = 0; while(n > 1){ counter += n / 5;//计算1-n有多少个数能被5整除 n = n / 5;//计算5^n } return counter; }}
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