[LeetCode/LintCode] Factorial Trailing Zeros

Java_oldboy 發布于2019-08-14 15:14 / 675人閱讀

摘要：是的倍數，先找有多少個個，然后找多少個個，補上，然后多少個個，補上個個個

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，先找有多少個5（1個0），然后找多少個25（2個0），補上，然后多少個125（3個0），補上……

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5 
        //i = 25, count += 5（5個25）＝ 30; i = （1個）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;
    }
}

GPU云服務器云服務器 Zeros Trailing Factorial

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