【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
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: [−231, 231 − 1].
class Solution
{
public:
bool isUgly(int num)
{
if(num <= 0) return false;
while(num % 2 == 0) num/=2; //提取因子（如果除得进就一直除）
while(num % 3 == 0) num/=3;
while(num % 5 == 0) 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.
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.
思路：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.
/*
如：
k[0] = 1, 1*2 1*3 1*5
k[1] = 2, 2*2 1*3 1*5
k[2] = 3, 2*2 2*3 1*5
*/
#include <algorithm>
class Solution
{
public:
int nthUglyNumber(int n)
{
if(n <= 0) return 0; //表示不存在
vector<int> k(n); //用于存各丑数
k[0] = 1;
int p2 = 0, p3 = 0, p5 = 0; //对需要乘因子2,3,5的数的指针
for(int i = 1; i<n; i++) //i = 1~n-1
{
k[i] = min(k[p2]*2, min(k[p3]*3, k[p5]*5));
if(k[i] == k[p2]*2) p2++; //只要相等就要移动，如6，需同时移动2和3的指针
if(k[i] == k[p3]*3) p3++;
if(k[i] == k[p5]*5) p5++;
}
return k[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.
Example:
Input: n = 12, primes = [2,7,13,19]Output: 32
Explanation: [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 is a super ugly number for any given primes.
-
The given numbers in primes are in ascending order.
-
0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
-
The nth super ugly number is guaranteed to fit in a 32-bit signed integer.
//类似ugly number II
#include <climits>
class Solution
{
public:
int nthSuperUglyNumber(int n, vector<int>& primes)
{
if(n <= 0) return 0; //表示不存在丑数
int k = primes.size();
vector<int> index(k, 0); //用于指向要乘某因子的丑数,初始化为0
vector<int> ugly(n);
ugly[0] = 1;
for(int i = 1; i<n; i++) //i=1~n-1
{
int temp = INT_MAX; //初始化为最大数,初始化为ugly[index[0]] * primes[0]较好
for(int j = 0; j<k; j++) temp = min(temp, ugly[index[j]] * primes[j]);
for(int j = 0; j<k; j++) //如果选取的是之前某个丑数乘某因子作为下个丑数，则该因子指针移动到下一个位置（否则下次乘出来还是最小的）
{
if(temp == ugly[index[j]] * primes[j]) index[j]++;
}
ugly[i] = temp;
}
return ugly[n-1];
}
};