880 Decoded String at Index
1. Question
An encoded stringS
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 stringS
, and an indexK
, find and return theK
-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 digits2
through9
.S
starts with a letter.1 <= K <= 10^9
The decoded string is guaranteed to have less than
2^63
letters.
2. Implementation
(1) Math
思路: 一看下最简单的做法就是解析String, 然后找出第K个位置的character。但这种做法其实不需要,因为我们只关心第K个位置的character,我们只要知道解析后的string的总长度,然后从后往前扫一遍输入的String, 当K等于0, 而当前位置的character不是digit,则我们找到第K个character
class Solution {
public String decodeAtIndex(String S, int K) {
long len = 0;
for (char c : S.toCharArray()) {
if (Character.isDigit(c)) {
len *= (c - '0');
}
else {
++len;
}
}
for (int i = S.length() - 1; i >= 0; i--) {
K %= len;
char c = S.charAt(i);
if (K == 0 && c >= 'a' && c <= 'z') {
return "" + c;
}
if (Character.isDigit(c)) {
len /= (c - '0');
}
else {
--len;
}
}
return "";
}
}
3. Time & Space Complexity
(1) Math: 时间复杂度O(n), 空间复杂度O(1)
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