Decoded String at Index (M)
題目
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-1more 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 <= 100Swill only contain lowercase letters and digits2through9.Sstarts with a letter.1 <= K <= 10^9- It's guaranteed that
Kis less than or equal to the length of the decoded string. - The decoded string is guaranteed to have less than
2^63letters.
題意
將給定字串按照指定規則展開后，取指定位置的字符，
思路
參考[LeetCode] 880. Decoded String at Index 在位置坐標處解碼字串，
代碼實作
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 "";
}
}
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