[leetcode-394-Decode String]

最新推荐文章于 2018-04-24 20:41:53 发布

weixin_34352005

最新推荐文章于 2018-04-24 20:41:53 发布

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原文链接：http://www.cnblogs.com/hellowooorld/p/7650864.html

本文介绍了解码特定格式字符串的算法实现。该算法通过递归方式处理输入字符串中的数字与方括号，返回解码后的结果。例如，输入3[a2[c]]将返回accaccacc。

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".

思路：

定义一个函数，参数为i和字符串s，代表从字符串的下标i开始处理，返回处理后的结果。

 string decodeString(const string& s, int& i)
 {
        string res;        
        while (i < s.length() && s[i] != ']')
　　　　 {
            if (!isdigit(s[i]))
                res += s[i++];
            else
　　　　　　  {
                int n = 0;
                while (i < s.length() && isdigit(s[i]))
                    n = n * 10 + s[i++] - '0';                    
                i++; // '['
                string t = decodeString(s, i);
                i++; // ']'                
                while (n-- > 0)
                    res += t;
            }
        }        
        return res;
    }

    string decodeString(string s)
    {
        int i = 0;
        return decodeString(s, i);
    }