Factorial Trailing Zeroes
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Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
思路:有多少个2与5相乘,就有多少个0,由于2远多于5,就是算有多少个5。
class Solution {public: int trailingZeroes(int n) { int len = 0; while(n >= 5) { len = len + n/5; n = n/5; } return len; }};
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