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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.
cracking interview原题,2*5可以构成一个10,而5的个数少于2的个数,因此只需找出多少的5即可。
public class Solution {
public int trailingZeroes(int n) {
int num = 0;
while (n >= 5 ) {
num += n / 5;
n = n / 5;
}
return num;
}
}
172. Factorial Trailing Zeroes
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原文地址:http://www.cnblogs.com/shini/p/4421248.html