LeetCode题解:Factorial Trailing Zeroes
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Given an integer n, return the number of trailing zeroes in n!.
题意:求n阶乘结果中有多少0
解决思路:阶乘要出现0必然能分解为5*2,2的数量大于5的数量,所以每一个5必然有一个2匹配,所以计算5的个数即可
代码:
public class Solution { public int trailingZeroes(int n) { int count = 0; while(n > 0){ n /= 5; count += n; } return count; }}
0 0
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