【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: [−2 31 , 2 31 − 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
n
th
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 ≤ 10 6 , 0 < primes[i] < 1000.
-
The n th 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
];
}
};