方法一:官方答案
class Solution { int ptr; public String decodeString(String s) { LinkedList<String> stk = new LinkedList<String>(); ptr = 0; while (ptr < s.length()) { char cur = s.charAt(ptr); if (Character.isDigit(cur)) { // 获取一个数字并进栈 String digits = getDigits(s); stk.addLast(digits); } else if (Character.isLetter(cur) || cur == '[') { // 获取一个字母并进栈 stk.addLast(String.valueOf(s.charAt(ptr++))); } else { ++ptr; LinkedList<String> sub = new LinkedList<String>(); while (!"[".equals(stk.peekLast())) { sub.addLast(stk.removeLast()); } Collections.reverse(sub); // 左括号出栈 stk.removeLast(); // 此时栈顶为当前 sub 对应的字符串应该出现的次数 int repTime = Integer.parseInt(stk.removeLast()); StringBuffer t = new StringBuffer(); String o = getString(sub); // 构造字符串 while (repTime-- > 0) { t.append(o); } // 将构造好的字符串入栈 stk.addLast(t.toString()); } } return getString(stk); } public String getDigits(String s) { StringBuffer ret = new StringBuffer(); while (Character.isDigit(s.charAt(ptr))) { ret.append(s.charAt(ptr++)); } return ret.toString(); } public String getString(LinkedList<String> v) { StringBuffer ret = new StringBuffer(); for (String s : v) { ret.append(s); } return ret.toString(); } }