[leetcode]394. Decode String

最新推荐文章于 2024-06-19 18:22:32 发布

xiaocong1990

最新推荐文章于 2024-06-19 18:22:32 发布

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本文介绍了解码字符串的算法实现，该算法通过栈来处理输入字符串中的数字和括号，重复括号内的字符串指定次数，最终返回解码后的完整字符串。文章提供了一个C++示例，演示如何处理如3[a]2[bc]这样的输入，输出aaabcbc。

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题目链接：https://leetcode.com/problems/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 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".

class Solution{
public:
    string decodeString(string s)
    {
        string ans;
        stack<string> S;
        stack<int> countStack;
        int i=0;
        while(i<s.length())
        {
            if(s[i]<='9' && s[i]>='0')
            {
                int count=0;
                while(s[i]<='9' && s[i]>='0')
                {
                    count=count*10+(s[i]-'0');
                    i++;
                }
                countStack.push(count);
            }
            else if (s[i]=='[')
            {
                S.push(ans);
                ans="";
                i++;
            }
            else if(s[i]==']')
            {
                string tmp=S.top();
                S.pop();
                int cnt=countStack.top();
                countStack.pop();
                for(int j=0;j<cnt;j++)
                    tmp+=ans;
                ans=tmp;
                i++;
            }
            else
            {
                ans+=s[i];
                i++;
            }
        }
        return ans;
    }
};