Factorial Trailing Zeroes
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Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
思路:如果要知道多少个0,就需要找到他们的约数中,有多少个2和5,由于2的个数必多于5,所以,找到5的个数即可
思路1:最笨的方法,暴力O(n),从2遍历到n,找到每个数中5的个数,累计求和
思路2:参考网上思路,给定一个数n,从1至n,仅含有一个5的数的个数为n/5,含有2个5的个数为n/25,因此,每次除以5的个数累计求和即可
代码:
class Solution {
public:
int trailingZeroes(int n) {
int res = 0;
while(n){
res += n/5;
n /= 5;
}
return res;
}
};
0 0
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