forked from fishercoder1534/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 1
/
_788.java
62 lines (55 loc) · 1.65 KB
/
_788.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
package com.fishercoder.solutions;
import java.util.HashMap;
import java.util.Map;
/**
* 788. Rotated Digits
*
* X is a good number if after rotating each digit individually by 180 degrees,
* we get a valid number that is different from X.
* A number is valid if each digit remains a digit after rotation.
* 0, 1, and 8 rotate to themselves;
* 2 and 5 rotate to each other;
* 6 and 9 rotate to each other,
* and the rest of the numbers do not rotate to any other number.
Now given a positive number N, how many numbers X from 1 to N are good?
Example:
Input: 10
Output: 4
Explanation:
There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.
Note: N will be in range [1, 10000].
*/
public class _788 {
public static class Solution1 {
public int rotatedDigits(int N) {
int count = 0;
Map<Character, String> map = new HashMap<>();
map.put('0', "0");
map.put('1', "1");
map.put('8', "8");
map.put('2', "5");
map.put('5', "2");
map.put('6', "9");
map.put('9', "6");
for (int i = 1; i <= N; i++) {
if (isRotatedNumber(i, map)) {
count++;
}
}
return count;
}
private boolean isRotatedNumber(int num, Map<Character, String> map) {
String originalNum = String.valueOf(num);
StringBuilder sb = new StringBuilder();
for (char c : String.valueOf(num).toCharArray()) {
if (!map.containsKey(c)) {
return false;
} else {
sb.append(map.get(c));
}
}
return !originalNum.equals(sb.toString());
}
}
}