172. 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.
链接: http://leetcode.com/problems/factorial-trailing-zeroes/
题解:
求n!里有多少个0。其实主要就是看有多少个5,有多少个5就有多少个0,这样我们就可以用一个while循环来搞定。
Time Complexity - O(logn), Space Complexity - O(1)
public class Solution { public int trailingZeroes(int n) { if(n <= 0) return 0; int res = 0; while(n > 0) { res += n / 5; n /= 5; } return res; } }
二刷:
找到在n里有多少个5,就可以得到结果。
Java:
Time Complexity - O(logn), Space Complexity - O(1)
public class Solution { public int trailingZeroes(int n) { if (n < 5) { return 0; } int count = 0; while (n > 0) { count += n / 5; n /= 5; } return count; } }
三刷:
题目可以转化为,求n!中含有多少个5。如何计算一个数里面有多少个5呢?我们可以用以下公式:
count = n / 5 + n / 25 + n / 125 + ....
就是用n除5来取得一开始的基数,当遇到5的倍数的时候,我们也要作相应的增加, 转换为循环的话我们可以先计算单个5的个数 n / 5,然后 n /= 5来计算 25的个数,然后再继续。最后返回count.
Java:
Time Complexity - O(logn), Space Complexity - O(1)
public class Solution { public int trailingZeroes(int n) { if (n < 5) { return 0; } int count = 0; while (n > 0) { count += n / 5; n /= 5; } return count; } }
Reference:
https://leetcode.com/discuss/62976/my-python-solution-in-o-1-space-o-logn-time
https://leetcode.com/discuss/19847/simple-c-c-solution-with-detailed-explaination
http://15838341661-139-com.iteye.com/blog/1883889