問題描述:
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.
源碼:
這題直接乘鐵不行,萬一n太大咋辦。於是我就想統計2和5得個數。結果還是超時了,最後兩個沒過。
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;
}
};
之後翻了一下discuss,意思大概就是5個個數肯定比2多,你想想小於n的2^i和5^j,j肯定小於2。所以只要求出5的個數即可。仍需注意的一點就是,像 25,125,這樣的不只含有一個5的數字需要考慮進去,時間100%,空間100%:
class Solution {
public:
int trailingZeroes(int n) {
long long int count=0;
while(n>=1){
count += n/5;
n = n/5;
}
return count;
}
};