题解 | #牛牛的字符串解码问题#

递归

class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param s string字符串
     * @return string字符串
     */
    pair<string, int> convertString(string s, int i) {
        string ret = "";
        string num = "";
        num += s[i];
        string substr = "";
        int idx = 0;
        for (int j = i + 1; j < s.length(); j++) {
            if (s[j] <= '9' && s[j] >= '0') num += s[j];
            else if (s[j] == '[') {
                for (int k = j + 1; k < s.length(); k++) {
                    if (s[k] >= 'a' && s[k] <= 'z') substr += s[k];
                    else if (s[k] == ']') {
                        idx = k;
                        break;
                    }
                    else if (s[k]>='1'&&s[k]<='9'){
                        auto p = convertString(s, k);
                        substr+=p.first;
                        k = p.second;
                    }
                }
                break;
            }
        }
        int time = atoi(num.c_str());
        for (int i = 0; i < time; i++) {
            ret += substr;
        }
        return {ret, idx};
    }
    string decodeString(string s) {
        // write code here
        string ret = "";
        for (int i = 0; i < s.length(); i++) {
            if (s[i] >= 'a' && s[i] <= 'z') ret += s[i];
            if (s[i] >= '1' && s[i] <= '9') {
                auto p = convertString(s, i);
                ret += p.first;
                i = p.second;
            }
        }
        return ret;
    }
};