leetcode 394. Decode String-优快云博客

本文链接：https://blog.youkuaiyun.com/liqinzhe11/article/details/78923302

本文介绍了一种用于解码特定格式字符串的算法。该算法能够处理形如 k[encoded_string] 的编码规则，其中 encoded_string 会被重复 k 次。通过递归地查找并替换最内层的编码字符串，实现了对复杂嵌套结构的有效解码。

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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 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".

思路就是从最里层的换起，所以找第一个']'。

class Solution {
public:
    string helper(string s, int i) //找到第一个']'然后往前找'['和'数字'、然后更改返回
    {
        int j = i - 1;
        while (s[j] != '[')    j--;
        int k = j - 1;
        while (s[k] >= '0' && s[k] <= '9') k--;
        string ret = s.substr(0, k + 1);
        int num = atoi(s.substr(k + 1, j - k - 1).c_str());
        for (int h = 0; h < num; h++)
            ret  = ret + s.substr(j + 1, i - j - 1);
        ret = ret + s.substr(i + 1);
        return ret;
    }

    string decodeString(string s)
    {
        int pos = s.find(']');
        if (pos < 0 || pos >= s.size()) return s;
        return decodeString(helper(s, pos));
    }
};