LeetCode : 394. Decode String 解码字符串重复复制子串

最新推荐文章于 2024-08-11 22:02:20 发布

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最新推荐文章于 2024-08-11 22:02:20 发布

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本文链接：https://blog.youkuaiyun.com/qq_16234613/article/details/90133191

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本文介绍了一种使用栈解决字符串解码问题的算法。通过解析如'3[a2[c]]'等编码格式，实现字符串的正确解码。文章详细解释了算法的实现过程，并提供了完整的代码示例。

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试题
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”.

代码
思路是利用栈，当出现]时说明要开始进行复制操作了。
"3[a2[c]]"为例：
将3，[，a，2，[，c，分别存入栈，
然后出现]，将2，[，c弹出，进行复制操作，复制完后压入栈。

class Solution {
    public String decodeString(String s) {
        if(s==null || s.length()==0) return "";
        Stack<String> de = new Stack<>();
        StringBuilder tmp = new StringBuilder();
        StringBuilder sss = new StringBuilder();
        StringBuilder merge = new StringBuilder();
        boolean flag = true;
        
        for(char c : s.toCharArray()){
            if(c=='['){
                de.push(tmp.toString());
                tmp.delete(0,tmp.length());
                de.push(String.valueOf(c));
            }else if(c==']'){
                de.push(tmp.toString());
                tmp.delete(0,tmp.length());
                sss.delete(0,sss.length());
                merge.delete(0,merge.length());
                
                while( !de.peek().equals("[") ){
                    sss.insert(0,de.pop());
                }
                de.pop();
                String re = de.pop();
                
                for(int i=0; i<Integer.valueOf(re); i++){
                    merge.append(sss);
                }
                de.push(merge.toString());
                
            }
            else{
                if(tmp.length()==0){
                    tmp.append(c);
                    if(c>='0' && c<='9'){
                        flag = true;
                    }else if(c>='a' && c<='z'){
                        flag = false;
                    }
                }else if( (flag&&(c>='0' && c<='9')) || (!flag&&(c>='a' && c<='z')) ){
                    tmp.append(c);
                }else{
                    de.push(tmp.toString());
                    tmp.delete(0,tmp.length());
                    tmp.append(c);
                    if(c>='0' && c<='9'){
                        flag = true;
                    }else if(c>='a' && c<='z'){
                        flag = false;
                    }
                }
            }
        }
        
        StringBuilder out = new StringBuilder();
        while(!de.isEmpty()){
            out.insert(0,de.pop());
        }
        if(tmp.length()!=0) out.append(tmp);
        return out.toString();
    }
}