leetcode 172. Factorial Trailing Zeroes
2016-06-02 15:33
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题目
Given an integer n, return the number of trailing zeroes in n!.Note: Your solution should be in logarithmic time complexity.
解1
public class Solution { public int trailingZeroes(int n) { //计算包含的2和5组成的pair的个数就可以了,一开始想错了,还算了包含的10的个数。 //因为5的个数比2少,所以2和5组成的pair的个数由5的个数决定。 //观察15! = 有3个5(来自其中的5, 10, 15), 所以计算n/5就可以。 //但是25! = 有6个5(有5个5来自其中的5, 10, 15, 20, 25, 另外还有1个5来自25=(5*5)的另外一个5), //所以除了计算n/5, 还要计算n/5/5, n/5/5/5, n/5/5/5/5, ..., n/5/5/5,,,/5直到商为0。 int count_five = 0; while ( n > 0) { int k = n / 5; count_five += k; n = k; } return count_five; } }
解2(超时)
public class Solution { public int trailingZeroes(int n) { if(n<1) return 0; int count_five=0; for(int i=n;i>=1;i--){ if(i%5==0){ //System.out.println(i); int N=i; while(N%5==0){ count_five++; N=N/5; } } } return count_five; } }
参考链接
参考链接:http://blog.csdn.net/feliciafay/article/details/42336835相关文章推荐
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