【Leetcode】Factorial Trailing Zeroes

【题目】

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

Note: Your solution should be in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

【思路】

10 is the product of 2 and 5. In n!, we need to know how many 2 and 5, and the number of zeros is the minimum of the number of 2 and the number of 5.

Since multiple of 2 is more than multiple of 5, the number of zeros is dominant by the number of 5.

Here we expand

  2147483647!=2 * 3 * ...* 5 ... *10 ... 15* ... * 25 ... * 50 ... * 125 ... * 250...=2 * 3 * ...* 5 ... * (5^1*2)...(5^1*3)...*(5^2*1)...*(5^2*2)...*(5^3*1)...*(5^3*2)... (Equation 1)

We just count the number of 5 in Equation 1.

Multiple of 5 provides one 5, multiple of 25 provides two 5 and so on.

Note the duplication: multiple of 25 is also multiple of 5, so multiple of 25 only provides one extra 5.

Here is the basic solution:

return n/5 + n/25 + n/125 + n/625 + n/3125+...;

You can easily rewrite it to a loop.

Because the trailing zeros is only related to the number of 5 in n!, we can calculate this by the code below

【代码】

  public int trailingZeroes(int n) {    int rs = 0;    while (n != 0) {        rs += (n / 5);        n /= 5;    }    return rs;}

int trailingZeroes(int n) {    return (n < 5  ? 0 : (n/5 + trailingZeroes(n/5)));}