394.Decode String

Tags: Medium Stack

Link: https://leetcode.com/problems/decode-string/

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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].

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Examples:

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

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Answer:

//不用stack的递归实现
class Solution {
public:
    string decoding(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 = decoding(s, i);
                i++; // ']'

                while (n-- > 0)
                    res += t;
            }
        }

        return res;
    }

    string decodeString(string s) {
        int i = 0;

        return decoding(s, i);
    }
};

//使用stack，但是运行速度很慢
class Solution {
public:
    string decodeString(string s) {
        string res = "";
        stack<int> num;
        stack<string> alphaString;
        int n = 0;

        for(auto c : s)
        {
            if(isdigit(c)) n = n * 10 + (c - '0'); //检测到数字
            else if(c == '['){ //检测到‘[’，把数字推入num，n归零
                num.push(n);
                n = 0;
                alphaString.push(res);
                res = "";
            }
            else if(isalpha(c)) res.push_back(c); //检测到字母，在遇到下一个']'之前都加入到res
            else{ //检测到']'
                string tmp = res;
                for(int i= 0; i < num.top() - 1; ++i) res += tmp;
                res = alphaString.top() + res;
                num.pop();
                alphaString.pop();
            }
        }

        return res;
    }
};