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‎README.adoc‎

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@@ -6184,13 +6184,13 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
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//|{doc_base_url}/0879-profitable-schemes.adoc[题解]
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//|Hard
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//|
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//
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//|{counter:codes}
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//|{leetcode_base_url}/decoded-string-at-index/[880. Decoded String at Index^]
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//|{source_base_url}/_0880_DecodedStringAtIndex.java[Java]
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//|{doc_base_url}/0880-decoded-string-at-index.adoc[题解]
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//|Medium
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//|
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|{counter:codes}
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|{leetcode_base_url}/decoded-string-at-index/[880. Decoded String at Index^]
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|{source_base_url}/_0880_DecodedStringAtIndex.java[Java]
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|{doc_base_url}/0880-decoded-string-at-index.adoc[题解]
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|Medium
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|
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|{counter:codes}
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|{leetcode_base_url}/boats-to-save-people/[881. Boats to Save People^]

‎docs/0880-decoded-string-at-index.adoc‎

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[#0880-decoded-string-at-index]
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= 880. Decoded String at Index
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= 880. 索引处的解码字符串
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{leetcode}/problems/decoded-string-at-index/[LeetCode - Decoded String at Index^]
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https://leetcode.cn/problems/decoded-string-at-index/[LeetCode - 880. 索引处的解码字符串^]
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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:
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给定一个编码字符串 `s` 。请你找出 __ *解码字符串* 并将其写入磁带。解码时，从编码字符串中 *每次读取一个字符*，并采取以下步骤：
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* 如果所读的字符是字母，则将该字母写在磁带上。
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* 如果所读的字符是数字（例如 `d`），则整个当前磁带总共会被重复写 `d-1` 次。
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* If the character read is a letter, that letter is written onto the tape.
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* If the character read is a digit (say `d`), the entire current tape is repeatedly written `d-1` more times in total.
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现在，对于给定的编码字符串 `s` 和索引 `k`，查找并返回解码字符串中的第 `k` 个字母。
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*示例 1：*
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Now for some encoded string `S`, and an index `K`, find and return the `K`-th letter (1 indexed) in the decoded string.
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....
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输入：s = "leet2code3", k = 10
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输出："o"
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解释：
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解码后的字符串为 "leetleetcodeleetleetcodeleetleetcode"。
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字符串中的第 10 个字母是 "o"。
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....
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*示例 2：*
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....
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输入：s = "ha22", k = 5
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输出："h"
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解释：
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解码后的字符串为 "hahahaha"。第 5 个字母是 "h"。
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....
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*Example 1:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* S = "leet2code3", K = 10
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*Output:* "o"
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*Explanation:*
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The decoded string is "leetleetcodeleetleetcodeleetleetcode".
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The 10th letter in the string is "o".
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----
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*Example 2:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* S = "ha22", K = 5
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*Output:* "h"
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*Explanation:*
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The decoded string is "hahahaha". The 5th letter is "h".
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----
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*Example 3:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* S = "a2345678999999999999999", K = 1
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*Output:* "a"
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*Explanation:*
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The decoded string is "a" repeated 8301530446056247680 times. The 1st letter is "a".
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----
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*Note:*
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* `2 <= S.length <= 100`
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* `S` will only contain lowercase letters and digits `2` through `9`.
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* `S` starts with a letter.
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* `1 <= K <= 10^9`
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* The decoded string is guaranteed to have less than `2^63` letters.
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*示例 3：*
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....
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输入：s = "a2345678999999999999999", k = 1
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输出："a"
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解释：
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解码后的字符串为 "a" 重复 8301530446056247680 次。第 1 个字母是 "a"。
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....
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*提示：*
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* `2 \<= s.length \<= 100`
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* `s` 只包含小写字母与数字 `2` 到 `9` 。
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* `s` 以字母开头。
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* `1 \<= k \<= 10^9^`
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* 题目保证 `k` 小于或等于解码字符串的长度。
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* 解码后的字符串保证少于 `2^63^` 个字母。
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== 思路分析
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逆向思维，正向求解遇到超大数字会堆栈溢出。需要逆向处理，从后向前，逐步把 `k` 值减小。
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[[src-0880]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0880_DecodedStringAtIndex.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0880_DecodedStringAtIndex_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/decoded-string-at-index/solutions/3548/suo-yin-chu-de-jie-ma-zi-fu-chuan-by-leetcode/[880. 索引处的解码字符串 - 官方题解^]
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. https://leetcode.cn/problems/decoded-string-at-index/solutions/111640/xian-ji-suan-zi-chuan-zhan-kai-zong-chang-du-zai-y/[880. 索引处的解码字符串 - 先計算字串展開總長度，再由原編碼字串由後向前作等價裁減^]
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. https://leetcode.cn/problems/decoded-string-at-index/solutions/2773275/880-suo-yin-chu-de-jie-ma-zi-fu-chuan-by-ft95/[880. 索引处的解码字符串 - 无题^]

‎docs/index.adoc‎

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// include::0879-profitable-schemes.adoc[leveloffset=+1]
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// include::0880-decoded-string-at-index.adoc[leveloffset=+1]
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include::0880-decoded-string-at-index.adoc[leveloffset=+1]
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‎logbook/202601.adoc‎

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|{doc_base_url}/0877-stone-game.adoc[题解]
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|✅ 数学题或者动态规划。双方都是聪明人，每次都选剩余石堆中的最多数，那么先选的人肯定获胜。
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|{counter:codes}
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|{leetcode_base_url}/decoded-string-at-index/[880. Decoded String at Index^]
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|{doc_base_url}/0880-decoded-string-at-index.adoc[题解]
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|⭕️ 栈。逆向思维，正向求解遇到超大数字会堆栈溢出。需要逆向处理，从后向前，逐步把 `k` 值减小。
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|===
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‎logbook/corrections.adoc‎

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|{doc_base_url}/0874-walking-robot-simulation.adoc[题解]
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|⭕️
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|{counter:codes}
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|{leetcode_base_url}/decoded-string-at-index/[880. Decoded String at Index^]
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|{doc_base_url}/0880-decoded-string-at-index.adoc[题解]
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|⭕️
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|===
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截止目前，本轮练习一共完成 {corrections} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0880_DecodedStringAtIndex.java‎

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package com.diguage.algo.leetcode;
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public class _0880_DecodedStringAtIndex {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2026-08-22 17:19:27
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*/
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public String decodeAtIndex(String s, int k) {
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long[] len = new long[s.length()];
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len[0] = 1L;
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for (int i = 1; i < s.length(); i++) {
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char c = s.charAt(i);
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if ('2' <= c && c <= '9') {
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len[i] = len[i - 1] * (c - '0');
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} else {
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len[i] = len[i - 1] + 1;
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}
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}
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for (int i = s.length() - 1; i >= 0; i--) {
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char c = s.charAt(i);
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k = (int) (k % len[i]);
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if (k == 0 && Character.isLetter(c)) {
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return String.valueOf(c);
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}
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}
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return null;
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}
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// end::answer[]
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static void main() {
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new _0880_DecodedStringAtIndex()
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// .decodeAtIndex("ha22", 5);
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// .decodeAtIndex("leet2code3", 10);
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.decodeAtIndex("y959q969u3hb22odq595", 222280369);
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}
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}

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