题目描述:

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

**Note: **Your solution should be in logarithmic time complexity.

统计结果中因子5的个数.

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class Solution {
public:
    int trailingZeroes(int n) {
        int ret = 0;
        while(n > 0){
            ret += n / 5; // 实际是计算比n小的最大的5的整数倍
            n /= 5; // 实际是计算比n小的最大的5的整数倍除以5
        }
        return ret;
    }
};