Leetcode--394. Decode String 递归

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本文深入探讨了解码字符串算法的实现细节，通过递归方法解析并还原由数字和方括号组成的编码字符串，如3[a]2[bc]转换为aaabcbc。文章提供了完整的C++代码示例，详细解释了如何处理重复次数和内部字符串，为读者提供了一个清晰的算法理解路径。

394. Decode String

Given an encoded string, return its 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 decode(string s, int& i) {
		string result;
		int count;
		while (i < s.size()) {
			if (isdigit(s[i])) {
				int beg = i;
				i++;
				while (i < s.size() && isdigit(s[i])) {
					i++;
				}
				count = stoi(s.substr(beg, i - beg));
			}
			else if (s[i] == '[') {
				i++;
				string tmp = decode(s, i);
				while (count--)
				{
					result = result + tmp;
				}
			}
			else {
				if (isalpha(s[i])) {
					result = result + s[i];
					i++;
				}
				else {
					i++;
					return result;
				}
			}
			
		}

		return result;
	}
	string decodeString(string s) {
		int i = 0;
		string res =decode(s, i);
		return res;
	}

};
int main() {
	Solution sol;
	cout<<sol.decodeString("3[a]2[b4[F]c]");
}