|
| 1 | +/** |
| 2 | + * An encoded string S is given. To find and write the decoded string to a |
| 3 | + * tape, the encoded string is read one character at a time and the following |
| 4 | + * steps are taken: |
| 5 | + * |
| 6 | + * If the character read is a letter, that letter is written onto the tape. |
| 7 | + * If the character read is a digit (say d), the entire current tape is |
| 8 | + * repeatedly written d-1 more times in total. |
| 9 | + * Now for some encoded string S, and an index K, find and return the K-th |
| 10 | + * letter (1 indexed) in the decoded string. |
| 11 | + * |
| 12 | + * Example 1: |
| 13 | + * Input: S = "leet2code3", K = 10 |
| 14 | + * Output: "o" |
| 15 | + * Explanation: |
| 16 | + * The decoded string is "leetleetcodeleetleetcodeleetleetcode". |
| 17 | + * The 10th letter in the string is "o". |
| 18 | + * |
| 19 | + * Example 2: |
| 20 | + * Input: S = "ha22", K = 5 |
| 21 | + * Output: "h" |
| 22 | + * Explanation: |
| 23 | + * The decoded string is "hahahaha". The 5th letter is "h". |
| 24 | + * |
| 25 | + * Example 3: |
| 26 | + * Input: S = "a2345678999999999999999", K = 1 |
| 27 | + * Output: "a" |
| 28 | + * Explanation: |
| 29 | + * The decoded string is "a" repeated 8301530446056247680 times. |
| 30 | + * The 1st letter is "a". |
| 31 | + * |
| 32 | + * Note: |
| 33 | + * 2 <= S.length <= 100 |
| 34 | + * S will only contain lowercase letters and digits 2 through 9. |
| 35 | + * S starts with a letter. |
| 36 | + * 1 <= K <= 10^9 |
| 37 | + * The decoded string is guaranteed to have less than 2^63 letters. |
| 38 | + */ |
| 39 | + |
| 40 | +publicclassDecodedStringAtIndex884 { |
| 41 | +publicStringdecodeAtIndex(StringS,intK) { |
| 42 | +char[]chars =S.toCharArray(); |
| 43 | +intN =chars.length; |
| 44 | +long[]lens =newlong[N]; |
| 45 | +longlen =0; |
| 46 | +for (inti=0;i<N;i++) { |
| 47 | +charc =chars[i]; |
| 48 | +if (c >='a' &&c <='z') { |
| 49 | +len++; |
| 50 | + }else { |
| 51 | +intre =Character.getNumericValue(c); |
| 52 | +len *=re; |
| 53 | + } |
| 54 | +lens[i] =len; |
| 55 | + } |
| 56 | + |
| 57 | +for (inti=N-1;i>=0;i--) { |
| 58 | +charc =chars[i]; |
| 59 | +if (c >='a' &&c <='z') { |
| 60 | +if (lens[i] ==K)returnCharacter.toString(c); |
| 61 | + }else { |
| 62 | +K %=lens[i-1]; |
| 63 | +if (K ==0) { |
| 64 | +intj =i-1; |
| 65 | +while (!(chars[j] >='a' &&chars[j] <='z')) { |
| 66 | +j--; |
| 67 | + } |
| 68 | +returnCharacter.toString(chars[j]); |
| 69 | + } |
| 70 | + } |
| 71 | + } |
| 72 | + |
| 73 | +returnCharacter.toString(chars[0]); |
| 74 | + } |
| 75 | + |
| 76 | +} |