Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
方法：求乘积中5的个数。
class Solution {
public:
int trailingZeroes(
int n) {
long num5 =
0;
long base =
5;
while(n/
base){
num5 += n/
base;
base *=
5;
}
return num5;
}
};
切忌不可用int类型定义，容易溢出。 防止溢出可以换种编程方式进行。
class Solution {
public:
int trailingZeroes(
int n) {
long num5 =
0;
while(n){
num5 += n/
5;
n/=
5;
}
return num5;
}
};
转载请注明原文地址: https://ju.6miu.com/read-666180.html
