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.
思路:实现很简单,关键的分析题意,要求n!末尾的0数,事实上可以求5和2的对数。而又因为出现5必定出现2,所以求5个数即可.
int trailingZeroes(int n) { int count = 0; n = n/5; while(n>0) { count+=n; n /=5; } return count; }
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