leetcode刷题，总结，记录 ，备忘 172

leetcode172 Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

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Special thanks to @ts for adding this problem and creating all test cases.

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求阶乘的尾数0的个数。首先想到的就是2x5，才会得0，又由于2的数量是肯定比5多的，所以只要计算5的数量即可，一开始我也比较天真，直接计算n / 5的结果，但是不行，因为还有25的情况，是5的2次方，需要多加个1，后面还有125，625等。所以使用循环每次除以5，将结果累加。

class Solution {
public:
int trailingZeroes(int n) {
int sum = 0;

while (n)
{
sum += n / 5;
n /= 5;
}

return sum;
}
};