[394]Decode String

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本文介绍了一种使用栈解决字符串解码问题的方法。该算法能够处理形如 k[encoded_string] 的编码规则，其中 encoded_string 被重复 k 次。文章提供了详细的解题思路及 C++ 实现代码。

【题目描述】

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".

【解题思路】

这道题有点跟以前数据结构书上的表达式求值类似，所以也可以通过遍历字符串用栈来实现。当然能用栈也就能用递归实现。

【代码】

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;
    }
};