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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.
给定一个整数n,让我们判断n的阶乘中末尾有几个零。我们只需要知道n的阶乘中有几个2和5的组合就可以,每个5前面都会有一个尾数为2的值,对于本身尾数为0的我们可以理解为它本身就是一个2和5的组合。这样我们只需要知道5的个数就可以了。代码如下:
Note: Your solution should be in logarithmic time complexity.
给定一个整数n,让我们判断n的阶乘中末尾有几个零。我们只需要知道n的阶乘中有几个2和5的组合就可以,每个5前面都会有一个尾数为2的值,对于本身尾数为0的我们可以理解为它本身就是一个2和5的组合。这样我们只需要知道5的个数就可以了。代码如下:
public class Solution { public int trailingZeroes(int n) { int count = 0; while(n > 0) { count += n / 5; n /= 5; } return count; } }
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