LeetCode-394. Decode String (JAVA)解码字符串

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".

解码字符串，本质是带有括号的计算器，思想同:224. Basic Calculator
LeetCode-224. Basic Calculator (JAVA)实现计算器

1. 没有中括号时正常运算，将结果存入res中

2. 遇到左括号时将res及括号之前的数字分别压栈

3. 遇到右括号时将取出左括号之前的数字（stack中取出），并加上左括号前的运算结果res（stack中取出），并拼接字符串：

stack中存入的是遇到左括号时①括号之前计算结果res，②括号之前的数字

public String decodeString(String s) {
	String res = "";
	// 记录'['之前的数字
	Stack countStack = new Stack<>();
	// 记录'['之前的运算结果
	Stack resStack = new Stack<>();
	int idx = 0;
	int curNum = 0;
	while (idx < s.length())

	{
		char ch = s.charAt(idx);
		if (Character.isDigit(ch)) {
			while (Character.isDigit(s.charAt(idx)))
				curNum = 10 * curNum 
					+ (s.charAt(idx++) - '0');
		} else if (ch == '[') {
			resStack.push(res);
			res = "";// 注意
			// 此push可以放在上面的while循环中
			countStack.push(curNum);
			curNum = 0;// 注意
			idx++;
			// 取出计算结果，和数字
		} else if (ch == ']') {
			StringBuilder temp = 
				new StringBuilder(resStack.pop());
			
			int repeatTimes = countStack.pop();
			for (int i = 0; i < repeatTimes; i++) {
				temp.append(res);
			}
			res = temp.toString();
			idx++;

			// 字母
		} else {
			res += s.charAt(idx++);
		}
	}
	return res;
}