394. Decode String**-优快云博客

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。通过使用两个栈分别记录重复次数与当前的解码结果，该算法可以高效地完成解码任务。

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

The encoding rule is: k[encoded_string], where the encoded_stringinside the square brackets is being repeated exactly k times. Note thatk 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".

public class Solution {
    public String decodeString(String s) {
        Stack<Integer> count = new Stack<>();
        Stack<String> result = new Stack<>();
        int i=0;
        result.push("");
        while(i<s.length()){
            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 times = count.pop();
                String str = result.pop();
                StringBuilder sb = new StringBuilder();
                for(int j=0;j<times;j++){
                    sb.append(str);
                }
                result.push(result.pop()+sb.toString());
            }else{
                result.push(result.pop()+ch);
            }
            i++;
            
        }
        return result.pop();
    }
}