Write an algorithm which computes the number of trailing zeros in n factorial.

**Example**

**11! = 39916800**, so the out should be 2

**Challenge**

O(log N) time

Solution: 只要找里面有多少个5, 乘出来就会有多少个0，因为2永远是够的

# of 5：n/5+n/25+n/(5^3)…..

public long trailingZeros(long n) { long base = 5;//important to use long type long count = 0; while (n >= base) { count += n / base; base *= 5; } return count; }