Google – Decompress String


Google – Decompress String
给一个string,比如abbaba4x[a]bb3x[abaa2x[bab]]。 4x[a]就表示aaaa, 2x[bab]就表示有两个连续的bab,babbab. 要求decompress 这个string。
[Solution]
[Solution #1]
recursion
碰到[]就递归整个括号里的东西,不管里面是否还有嵌套,也不管有几层嵌套。把递归回来的string整个append到stringbuilder,然后继续后面的。在碰到'['时需要找到与之对应的']',而不是在那之后的第一个close bracket!
[Solution #2]
Iterative
用一个stack来记录所有的Character, 一旦碰到close bracket ], 从stack里pop直到pop出open bracket。这样可以保证这两个括号肯定是一对并且在最里层。decompress之后把得到的结果再push回stack。继续后面的。
// recursive
class Solution {
  public String decompression(String s) {
    if (s == null || s.isEmpty()) {
      return s;
    }
 
    StringBuilder result = new StringBuilder();
 
    for (int i = 0; i < s.length(); i++) {
      char c = s.charAt(i);
      if (c >= 'a' && c <= 'z') {
        result.append(c);
      }
      else if (c >= '1' && c <= '9') {
        StringBuilder tmp = new StringBuilder();
        int xIdx = s.indexOf("x", i);
        int cnt = Integer.parseInt(s.substring(i, xIdx));
        i = xIdx + 2;
        int openLeft = 1;
        while (true) {
          if (s.charAt(i) == ']') {
            openLeft--;
            if (openLeft == 0) {
              break;
            }
            else {
              tmp.append(s.charAt(i));
            }
          }
          else if (s.charAt(i) == '[') {
            openLeft++;
            tmp.append(s.charAt(i));
          }
          else {
            tmp.append(s.charAt(i));
          }
          i++;
        }
 
        String sub = decompression(tmp.toString());
        for (int k = 0; k < cnt; k++) {
          result.append(sub);
        }
      }
    }
 
    return result.toString();
  }
}

// iterative stack
class Solution2 {
  public String decompression(String s) {
    if (s == null || s.isEmpty()) {
      return s;
    }
 
    Stack<Character> stack = new Stack<>();
    for (char c : s.toCharArray()) {
      if (c != ']') {
        stack.push(c);
      }
      else {
        List<Character> pattern = new LinkedList<>();
        while (true) {
          char top = stack.pop();
          if (top == '[') {
            break;
          }
          else {
            pattern.add(0, top);
          }
        }
        stack.pop();
        StringBuilder tmp = new StringBuilder();
        while (stack.peek() >= '0' && stack.peek() <= '9') {
          tmp.insert(0, stack.pop());
        }
        int cnt = Integer.parseInt(tmp.toString());
 
        for (int i = 0; i < cnt; i++) {
          for (char p : pattern) {
            stack.push(p);
          }
        }
      }
    }
 
    StringBuilder result = new StringBuilder();
    while (!stack.isEmpty()) {
      result.append(stack.pop());
    }
 
    return result.reverse().toString();
  }
}
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