394. Decode String

最新推荐文章于 2024-07-30 15:25:02 发布

BigFatSheep

最新推荐文章于 2024-07-30 15:25:02 发布

阅读量127

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode Medium

本文链接：https://blog.youkuaiyun.com/BigFatSheep/article/details/83059255

Leetcode 同时被 2 个专栏收录

385 篇文章

订阅专栏

Medium

195 篇文章

订阅专栏

本文介绍了解码特定格式的压缩字符串的算法，通过使用两个栈分别存储数字和字符串，实现复杂字符串的解码，适用于含有重复子串的字符串解压。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

Given an encoded string, return it’s decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won’t be input like 3a or 2[4].

Examples:

s = “3[a]2[bc]”, return “aaabcbc”.
s = “3[a2[c]]”, return “accaccacc”.
s = “2[abc]3[cd]ef”, return “abcabccdcdcdef”.

Problem URL

Solution

给一个string表示的压缩string，如有多个重复出现的字段使用[]扩起来。将它还原为原来的string。使用两个栈完成。

This problem is kind like recursion. We must know the deepest substring to get the outers. So we use two stack to store result and number count. For normal word, just follow behind res. For number, calculate it and store it in count. For “[”, it is the sign of a new recursion, we store res and clear it for next level. For “]”, it is sign of end recursion, pop count and res in stacks, use a string builder to build a decoded string.

Code

class Solution {
    public String decodeString(String s) {
        String res = "";
        Stack<Integer> countStack = new Stack<>();
        Stack<String> resStack = new Stack<>();
        int idx = 0;
        while (idx < s.length()) {
            if (Character.isDigit(s.charAt(idx))) {
                int count = 0;
                while (Character.isDigit(s.charAt(idx))) {
                    count = 10 * count + (s.charAt(idx) - '0');
                    idx++;
                }
                countStack.push(count);
            }
            else if (s.charAt(idx) == '[') {
                resStack.push(res);
                res = "";
                idx++;
            }
            else if (s.charAt(idx) == ']') {
                StringBuilder temp = new StringBuilder (resStack.pop());
                int repeatTimes = countStack.pop();
                for (int i = 0; i < repeatTimes; i++) {
                    temp.append(res);
                }
                res = temp.toString();
                idx++;
            }
            else {
                res += s.charAt(idx++);
            }
        }
        return res;
    }
}