leetcode 172. Factorial Trailing Zeroes
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Given an integer n, return the number of trailing zeroes in n!.
Example 1:
Input: 3 Output: 0 Explanation: 3! = 6, no trailing zero.
Example 2:
Input: 5 Output: 1 Explanation: 5! = 120, one trailing zero.
class Solution(object):
def trailingZeroes(self, n):
# 15!=3, 15/5=3
# 25!=6, 25/5=5+5^2|1
# count 5 number
if n < 5:
return 0
return n/5 + self.trailingZeroes(n/5)
思考这类问题还是要从上而下思考，先用递归思路去解，最后修改循环或者dp。
class Solution(object):
def trailingZeroes(self, n):
# 15!=3, 15/5=3
# 25!=6, 25/5=5+5^2|1
# count 5 number
ans = 0
while n >= 5:
ans += n/5
n = n/5
return ans
另外的解法：
Example Three
By given number 4617.
5^1 : 4617 ÷ 5 = 923.4, so we get 923 factors of 5
5^2 : 4617 ÷ 25 = 184.68, so we get 184 additional factors of 5
5^3 : 4617 ÷ 125 = 36.936, so we get 36 additional factors of 5
5^4 : 4617 ÷ 625 = 7.3872, so we get 7 additional factors of 5
5^5 : 4617 ÷ 3125 = 1.47744, so we get 1 more factor of 5
5^6 : 4617 ÷ 15625 = 0.295488, which is less than 1, so stop here.
Then 4617! has 923 + 184 + 36 + 7 + 1 = 1151 trailing zeroes.
C/C++ code
int trailingZeroes(int n) {
int result = 0;
for(long long i=5; n/i>0; i*=5){
result += (n/i);
}
return result;
}
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