leetcode 172: Factorial Trailing Zeroes

问题描述：

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

Note: Your solution should be in logarithmic time complexity.

思路：

问题要求 n! 的零尾个数，小数试验一下就知道，2的个数太多不用管，关键是求5和5的幂的个数。所以就求对于每一个k，n/(5^k)，k属于{k|5^k<=n,k>0}

但是在用int来保存 5^k的值时，在极限情况下会出现超出int范围的值，所以应该用unsigned int 或是 long int。后来发现 unsigned int 还是很不安全，只是比 int 多一倍，但乘幂一下子就会是 5 倍，仍然会出现溢出。

代码：

class Solution {
public:int trailingZeroes(int n) {int amount = 0;long power_five = 1;    //if not unsigned, may be exceed the int 2147483647 while(1){power_five *= 5;if(power_five > n) break;int addition = n / power_five;     //the number of "5's power" is the so important that it doesn't matter neglecting the 2.amount += addition;}return amount;}
};