[LeetCode] Factorial Trailing Zeroes 求階乘末尾零的個數
阿新 • • 發佈:2018-12-27
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.
這道題並沒有什麼難度,是讓求一個數的階乘末尾0的個數,也就是要找乘數中10的個數,而10可分解為2和5,而我們可知2的數量又遠大於5的數量,那麼此題即便為找出5的個數。仍需注意的一點就是,像25,125,這樣的不只含有一個5的數字需要考慮進去。程式碼如下:
C++ 解法一:
class Solution { public: int trailingZeroes(int n) { int res = 0; while (n) { res += n / 5; n /= 5; } return res; } };
Java 解法一:
public class Solution {
public int trailingZeroes(int n) {
int res = 0;
while (n > 0) {
res += n / 5;
n /= 5;
}
return res;
}
}
這題還有遞迴的解法,思路和上面完全一樣,寫法更簡潔了,一行搞定碉堡了。
C++ 解法二:
class Solution { public: int trailingZeroes(int n) { return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5); } };
Java 解法二:
public class Solution { public int trailingZeroes(int n) { return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5); } }
參考資料: