394. Decode String-优快云博客

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/72357606

本文介绍了一种用于解码特定格式字符串的算法实现。该算法能够处理包含数字与方括号的字符串，根据数字与括号内的子串重复指定次数来生成最终解码后的字符串。文章提供了详细的解码逻辑及Java实现代码。

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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".

这里需要分成4种情况考虑：

1、ch = 数字，向后遍历直到没有数字，把数字推入count堆栈

2、ch = [，string堆栈推入空字符串“”，开启新一篇

3、ch = ]， count出栈，stringbuilder.append() n 次

4、ch = 字母，result.push(result.pop() + ch)，这个设计很巧妙

代码如下：

public class Solution {
    public String decodeString(String s) {
        Stack<Integer> count = new Stack<Integer>();
        Stack<String> result = new Stack<String>();
        result.push("");
        int i = 0, len = s.length();
        while (i < len) {
            char ch = s.charAt(i); 
            if (ch >= '0' && ch <= '9') {
                int start = i;
                while (s.charAt(i + 1) >= '0' && s.charAt(i + 1) <= '9') i ++;
                count.push(Integer.parseInt(s.substring(start, i + 1)));
            } else if (ch == '[') {
                result.push("");
            } else if (ch == ']') {
                int n = count.pop();
                StringBuilder builder = new StringBuilder();
                String temp = result.pop();
                while (n > 0) {
                    builder.append(temp);
                    n --;
                }
                result.push(result.pop() + builder.toString());
            } else {
                result.push(result.pop() + ch);
            }
            i ++;
        }
        return result.pop();
    }
}