Leetcode #172 Factorial Trailing Zeroes
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Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
Difficulty: Easy.
The trailing zeros depends on the numbers of 5 in (0, n). Also, we should notice the size of int.
class Solution {public: int trailingZeroes(int n) { int ans = n/5; long long int num = 25; while(num<=n){ ans = ans + n/num; num = num * 5; } return ans; }};
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