Ugly Number
263. Ugly Number
Write a program to check whether a given number is an ugly number.
Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.
Example 1:
Input: 6
Output: true
Explanation: 6 = 2 × 3
Example 2:
Input: 8
Output: true
Explanation: 8 = 2 × 2 × 2
Example 3:
Input: 14
Output: false
Explanation: 14 is not ugly since it includes another prime factor 7.
Note:
- 1 is typically treated as an ugly number.
- Input is within the 32-bit signed integer range: [−2^31, 2^31 − 1].
题目：判断输入的数是否为丑数。
思路：分别对2,3,5进行取余判断，如果取余为0，则将num除以对应的数，判断最后剩下的值是否为1。
class Solution {
public:
bool isUgly(int num) {
if(num <= 0) return false;
for(int i = 2; i <=5; ++i){
while(num % i == 0){
num /= i;
}
}
return num == 1;
}
};
264. Ugly Number II
Write a program to find the n-th ugly number.
Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.
Example:
Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.
Note:
- 1 is typically treated as an ugly number.
- n does not exceed 1690.
题目：第n个丑数。
思路：参见C++ DP solution with short explanation.
We have an array k of first n ugly number. We only know, at the beginning, the first one, which is 1. Then
k[1] = min( k[0]x2, k[0]x3, k[0]x5). The answer is k[0]x2. So we move 2’s pointer to 1. Then we test:
k[2] = min( k[1]x2, k[0]x3, k[0]x5). And so on. Be careful about the cases such as 6, in which we need to forward both pointers of 2 and 3.
x here is multiplication.
class Solution {
public:
int nthUglyNumber(int n) {
if(n <= 0) return 0;
if(n == 1) return 1;
vector<int> k(n);
k[0] = 1;
int t2 = 0, t3 = 0, t5 = 0;
for(int i = 1; i < n; ++i){
k[i] = min(2*k[t2], min(3*k[t3], 5*k[t5]));
if(2*k[t2] == k[i]) ++t2;
if(3*k[t3] == k[i]) ++t3;
if(5*k[t5] == k[i]) ++t5;
}
return k[n-1];
}
};