Leetcode: Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

Naive方法：A simple method is to first calculate factorial of n, then count trailing 0s in the result (We can count trailing 0s by repeatedly dividing the factorial by 10 till the remainder is 0). 但这样做的显著缺点是：can cause overflow for a slightly bigger numbers as factorial of a number is a big number (See factorial of 20 given in above examples). 

转自GeeksforGeeks的想法：The idea is to consider prime factors of a factorial n. A trailing zero is always produced by prime factors 2 and 5. If we can count the number of 5s and 2s, our task is done. Consider the following examples.

n = 5: There is one 5 and 3 2s in prime factors of 5! (2 * 2 * 2 * 3 * 5). So count of trailing 0s is 1.

n = 11: There are two 5s and eight 2s in prime factors of 11! (2 ⁸ * 3⁴ * 5² * 7). So count of trailing 0s is 2.

我们会发现： the number of 2s in prime factors is always more than or equal to the number of 5s. So if we count 5s in prime factors, we are done. 

How to count total number of 5s in prime factors of n!? A simple way is to calculate floor(n/5). 

问题转化为求阶乘过程中质因子5的个数，但是要注意25能提供2个5,125能提供3个5....

所以，count= floor(n/5) + floor(n/25) + floor(n/125) + ....

public class Solution {
    public int trailingZeroes(int n) {
        int count = 0;
        for (int i=5; (n/i)>=1;) {
            count += n / i;
            n /= 5;
        }
        return count;
    }
}

最开始我的写法是：

// Function to return trailing 0s in factorial of n
int findTrailingZeros(int  n)
{
    // Initialize result
    int count = 0;
 
    // Keep dividing n by powers of 5 and update count
    for (int i=5; n/i>=1; i *= 5)
          count += n/i;
 
    return count;
}

在oj上提交会发现n =  1808548329时出错了，期望答案是452137076，实际答案是452137080

原因就是 i*5一直连乘时出现i = 5^14时，内存溢出(5^13 = 1220703125 < 2^31, but 5^14 = 6103515625 > 2^32)

Integer overflow之后会wrap around, 即Integer.MAX_VALUE + 1会成为Integer.MIN_VALUE, 详见Why Integer overflows wrap around

6103515625 wrap around之后 为正的1808548329-1 = 1808548328

原因是6103515625 % 2^32 = 1808548329 < 2 ^31，即 i 比32位Integer（共2^32）多出1808548329个数， 为 1808548328，又可以再进一次for 循环（本不应该进的）。所以答案偏大

解决办法：用除法代替乘法，用n / 5代替 i * 5，防止overflow，如最上面那段code所示