【LeetCode】172. Factorial Trailing Zeroes

LeetCode-127

Links:https://leetcode.com/problems/factorial-trailing-zeroes/

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

题目大意：求n！的末尾有多少个0…

思路：计算n！的质因数中5的个数。

因为：10 = 2*5（即只有一对2和5的时候才进位，又2比5个数多…）

class Solution {
public:
    int trailingZeroes(int n) {
        if(n<=4)return 0;
        else{
            int c = 0;
            while(n)
            {
                n/=5;
                c+=n;
            }
            return c;
        }
    }
};

class Solution {

public:

int trailingZeroes(int n) {

if(n<=4)return 0;

else{

int c = 0;

while(n)

{

n/=5;

c+=n;

}

return c;

}

};

【LeetCode】172. Factorial Trailing Zeroes

Tagged on: LeetCode