LeetCode 394. Decode String 题解

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本文介绍了解码特定格式字符串的方法，该格式使用k[encoded_string]表示encoded_string被重复k次。文章提供了详细的解题思路及C++代码实现，通过递归方式解析并构造最终的解码字符串。

394. Decode String

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 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 or2[4].

Examples:

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

解题思路：

请参考代码注释部分。

代码展示:

class Solution {
public:
    string dfs(string s,int &i)
    {
        string res;
        int l=s.length();
        //观察字符串的特点
        //如"xyz2[gf]3[a2[2[b5[cdf]]]]"
        //若开头不是数字，直接将字符加入结果
        //如果是数字，则首先得到字符串重复次数
        //然后对剩余的字符串递归调用dfs
        //注意到s[i]==']'时即退出循环
        //即退出内层方括号包括的字符串,返回结果到上一层
        while(i<l&&s[i]!=']')
        {
  
            if(!isdigit(s[i]))
            {
            	res+=s[i];
            	i++;
			}
			else
			{
				int cnt=0;
            	while(isdigit(s[i])&&i<l)
            	{
                	cnt*=10;
                	cnt+=s[i]-'0';
                	i++;
           	 	}
            	i++;
            	string tmp=dfs(s,i);
            	i++;
            	while(cnt--) res+=tmp;
			}
			
        }
        return res;
    }
    string decodeString(string s) {

        int n=s.size();
        string res;
        if(!n) return res;
        int i=0;
        return dfs(s,i);
    }
};