Leetcode NO.172 Factorial Trailing Zeroes
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本题题目要求如下:
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
应该说这算是leetcode里面最简单的一道题了。。基本就是我天朝小学数学的水平。。。。数多少个0其实本质上就是数有多少个2 * 5,比如10本身就贡献了2*5,像25,就贡献了2个5
而且需要注意的是,无论怎样,2肯定要比5多。。
比如1,2,3,4,5,6,7,8,9,10里面只有3个5,却有数不清的2,反正肯定大于3,所以这道题就简化为数1到n之间的数有多少个5
详细代码如下:
class Solution {public: int trailingZeroes(int n) { /* count the number of 5 */ int cnt = 0; while (n/5 >= 1) { n /= 5; cnt += n; } return cnt; }};
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