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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答案解释可以看 陈皓在leetcode Discuss上的答案：

int trailingZeroes(int n) {    int result = 0;    for (long long i = 5; n / i>0; i *= 5) {        result += (n / i);    }    return result;}