问题描述:
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.
Note: Your solution should be in logarithmic time complexity.
源码:
这题直接乘铁不行,万一n太大咋办。于是我就想统计2和5得个数。结果还是超时了,最后两个没过。
class Solution {
public:
int trailingZeroes(int n) {
int result = 0;
int count_2=0, count_5=0;
for(int i=2; i<=n; i++){
int tmp = i;
while(tmp % 10 == 0){
result++; tmp = tmp/10;
}
while(tmp % 2 == 0){
count_2++; tmp = tmp/2;
}
while(tmp % 5 == 0){
count_5++; tmp = tmp/5;
}
}
result += min(count_2, count_5);
return result;
}
};
之后翻了一下discuss,意思大概就是5个个数肯定比2多,你想想小于n的2^i和5^j,j肯定小于2。所以只要求出5的个数即可。仍需注意的一点就是,像 25,125,这样的不只含有一个5的数字需要考虑进去,时间100%,空间100%:
class Solution {
public:
int trailingZeroes(int n) {
long long int count=0;
while(n>=1){
count += n/5;
n = n/5;
}
return count;
}
};