-
Notifications
You must be signed in to change notification settings - Fork 94
/
Copy pathDecodedStringAtIndex884.java
76 lines (72 loc) · 2.24 KB
/
DecodedStringAtIndex884.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
63
64
65
66
67
68
69
70
71
72
73
74
75
76
/**
* An encoded string S is given. To find and write the decoded string to a
* tape, the encoded string is read one character at a time and the following
* steps are taken:
*
* If the character read is a letter, that letter is written onto the tape.
* If the character read is a digit (say d), the entire current tape is
* repeatedly written d-1 more times in total.
* Now for some encoded string S, and an index K, find and return the K-th
* letter (1 indexed) in the decoded string.
*
* Example 1:
* Input: S = "leet2code3", K = 10
* Output: "o"
* Explanation:
* The decoded string is "leetleetcodeleetleetcodeleetleetcode".
* The 10th letter in the string is "o".
*
* Example 2:
* Input: S = "ha22", K = 5
* Output: "h"
* Explanation:
* The decoded string is "hahahaha". The 5th letter is "h".
*
* Example 3:
* Input: S = "a2345678999999999999999", K = 1
* Output: "a"
* Explanation:
* The decoded string is "a" repeated 8301530446056247680 times.
* The 1st letter is "a".
*
* Note:
* 2 <= S.length <= 100
* S will only contain lowercase letters and digits 2 through 9.
* S starts with a letter.
* 1 <= K <= 10^9
* The decoded string is guaranteed to have less than 2^63 letters.
*/
public class DecodedStringAtIndex884 {
public String decodeAtIndex(String S, int K) {
char[] chars = S.toCharArray();
int N = chars.length;
long[] lens = new long[N];
long len = 0;
for (int i=0; i<N; i++) {
char c = chars[i];
if (c >= 'a' && c <= 'z') {
len++;
} else {
int re = Character.getNumericValue(c);
len *= re;
}
lens[i] = len;
}
for (int i=N-1; i>=0; i--) {
char c = chars[i];
if (c >= 'a' && c <= 'z') {
if (lens[i] == K) return Character.toString(c);
} else {
K %= lens[i-1];
if (K == 0) {
int j = i-1;
while (!(chars[j] >= 'a' && chars[j] <= 'z')) {
j--;
}
return Character.toString(chars[j]);
}
}
}
return Character.toString(chars[0]);
}
}