Decoded String at Index (M)
题目
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 writtend-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".
Constraints:
2 <= S.length <= 100
-
S
will only contain lowercase letters and digits2
through9
. -
S
starts with a letter. 1 <= K <= 10^9
- It's guaranteed that
K
is less than or equal to the length of the decoded string. - The decoded string is guaranteed to have less than
2^63
letters.
题意
将给定字符串按照指定规则展开后,取指定位置的字符。
思路
参考[LeetCode] 880. Decoded String at Index 在位置坐标处解码字符串。
代码实现
Java
class Solution {
public String decodeAtIndex(String S, int K) {
long size = 0;
int index = 0;
while (index < S.length() && size < K) {
char c = S.charAt(index);
size = Character.isDigit(c) ? size * (c - '0') : size + 1;
index++;
}
while (--index >= 0) {
char c = S.charAt(index);
if (Character.isDigit(c)) {
size /= c - '0';
K = (int)((K - 1) % size + 1);
} else {
if (K == size) {
return S.charAt(index) + "";
}
size--;
}
}
return "";
}
}