172. Factorial Trailing Zeroes
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1.Question
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
2.Codeclass Solution {public: int trailingZeroes(int n) { int count = 0; int factor = 5; for(int i = 1; i <= 13; i++) { count += n / factor; factor *= 5; } return count; }};3.Note
a. 该问题是求n! 末尾有几个零,经过分析后可以发现,n! 的分解因子中 5*2 才会产生0,然而2的个数肯定比5多,那么可以把该问题转换为“分解因子中有多少个5”。
b. 通过 n / factor (factor = 5, 25, 125 ,625...)可以求5个个数。然后是,int 型最大数字为4,294,967,295,则再循环中让factor不断变大的过程中不能溢出,5^13 = 1,220,703,125。则可以确定循环的次数为13次。
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- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
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