Leet Code OJ 172. Factorial Trailing Zeroes [Difficulty: Easy]
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题目:
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
翻译:
给定一个整数n,返回n!末尾的0的个数。
提示:你的解决方案应该运行在O(log(n))的时间复杂度。
分析:
我们知道要形成一个末尾的0,需要组合出2和5这两个质因子。然而在n!中,质因子2的个数永远大于5的个数。所以,问题转换为对n!计算5的质因子个数。
代码:
public class Solution { public int trailingZeroes(int n) { int s=0; while(n>=5){ n/=5; s+=n; } return s; }}
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