Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

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

 

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答案解释可以看 陈皓在leetcode Discuss上的答案：https://leetcode.com/discuss/19847/simple-c-c-solution-with-detailed-explaination

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