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[LeetCode/LintCode] Factorial Trailing Zeros

Java_oldboy / 724人閱讀

摘要：是的倍數(shù)，先找有多少個(gè)個(gè)，然后找多少個(gè)個(gè)，補(bǔ)上，然后多少個(gè)個(gè)，補(bǔ)上個(gè)個(gè)個(gè)

Problem

Write an algorithm which computes the number of trailing zeros in n factorial.

Challenge

11! = 39916800, so the output should be 2

Note

i是5的倍數(shù)，先找有多少個(gè)5（1個(gè)0），然后找多少個(gè)25（2個(gè)0），補(bǔ)上，然后多少個(gè)125（3個(gè)0），補(bǔ)上……

Solution 
class Solution {
    public long trailingZeros(long n) {
        long i = 5;
        long count = 0;
        while (i <= n) { 
        //n = 125, i = 5, count = 25; 25個(gè)5 
        //i = 25, count += 5（5個(gè)25）＝ 30; i = （1個(gè)）125, count  += 1 = 31; 
            count += n / i;
            i = i * 5;
        }
        return count;
    }
}
LeetCode version 
class Solution {
    public int trailingZeroes(int n) {
        int count = 0;
        while (n > 0) {
            count += n/5;
            n /= 5;
        }
        return count;
    }
}

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