Leetcode 172 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.
求阶乘的后缀0个数乘法中的零来源于10,10来源于2和5,在阶乘中,一个数的质因子出现一次5,那么必然有其他数的质因子出现若干次2
所以问题变为求解质因子5出现的次数,
n/5求出包含一个5的数字个数
n/25求出包含两个5的数字个数...以此类推
class Solution {public: int trailingZeroes(int n) { int res = 0; while(n) { res += n/5; n/=5; } return res; }};
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