leetcode 263|264|313. Ugly Number 1|2 313. Super 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

.
For example, 

6, 8

 are ugly while 

14

 is
not ugly since it includes another prime factor 

7

.

Note that 

1

 is typically treated as an ugly number.

class Solution {
public:
bool isUgly(int num)
{
if (num == 0) return false;
while (num % 2 == 0)
{
num = num/2;
}
while (num % 3 == 0)
{
num = num/3;
}
while (num % 5 == 0)
{
num = num/5;
}
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

.
For example, 

1, 2, 3, 4, 5, 6, 8, 9, 10, 12

is the sequence of the first 

10

 ugly
numbers.

Note that 

1

 is typically treated as an ugly number, and n does
not exceed 1690.
利用set自动排序的功能。

class Solution {
public:
int nthUglyNumber(int n)
{
int count = 0;
set<long long> se;
se.insert(1);
while (1)
{
long long key = *se.begin();
se.erase(key);
if ( ++count == n)
return key;
se.insert(key * 2);
se.insert(key * 3);
se.insert(key * 5);
}
return -1;
}
};

还有另外一种方法，val2 val3 val5里面中最小的那个就是下一个填进去的值，填进去之后，index后移，index2 就是 下一个*2的候选值的下标 

class Solution {
public:
int nthUglyNumber(int n)
{
vector<int> vc(n, 1);
int index2 = 0;
int index3 = 0;
int index5 = 0;
int val2 = 2;
int val3 = 3;
int val5 = 5;

for (int i = 1; i < n; i++)
{
int insert = min(val2, min(val3, val5));
vc[i] = insert;
if (insert == val2)
{
val2 = vc[++index2] * 2;
}
if (insert == val3)
{
val3 = vc[++index3] * 3;
}
if (insert == val5)
{
val5 = vc[++index5] * 5;
}
}
return vc[n - 1];
}
};
313. Super Ugly Number

Write a program to find the nth super ugly number.

Super ugly numbers are positive numbers whose all prime factors are in the given prime list 

primes

 of
size 

k

. For example, 

[1,
2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] 

is the sequence of the first 12 super ugly numbers given 

primes

 = 

[2,
7, 13, 19]

of size 4.

Note:

(1) 

1

 is a super ugly number for any given 

primes

.

(2) The given numbers in 

primes

 are in ascending order.

(3) 0 < 

k

 ≤ 100, 0 < 

n

 ≤
106, 0 < 

primes[i]

 <
1000.

(4) The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

方法和上面那个一模一样。

class Solution {
public:
int nthSuperUglyNumber(int n, vector<int>& primes)
{
vector<int> index(primes.size(), 0); //index[i] 存的是primes第i位应该和 index[i]相乘
vector<int> ret(n, 1);  //存结果数组
vector<int> val(primes);
for (int i = 1; i < n; i++)
{
ret[i] = *min_element(val.begin(), val.end());
for (int j = 0; j < val.size(); j++)
{
if (ret[i] == val[j])
{
val[j] = ret[++index[j]] * primes[j];
}
}
}
return ret[n - 1];
}
};