Go Java 算法之字符串解码示例详解

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();
    }
}

func decodeString(s string) string {
	var decode func(start int) (string, int)
	decode = func(start int) (str string, end int) {
		num:= 0
		for i := start; i < len(s); i++ {
			if isNumber(s[i]) {
				num = num*10 + int(s[i]-'0')
			} else if isLetter(s[i]) {
				str += string(s[i])
			} else if s[i] == '[' {
				item, index := decode(i + 1)
				for num != 0 {
					str += item
					num--
				}
				i = index
			} else if s[i] == ']' {
				end = i
				break
			}
		}
		return str, end
	}
	res, _ := decode(0)
	return res
}
func isLetter(u uint8) bool {
	return u >= 'A' && u <= 'Z' || u >= 'a' && u <= 'z'
}
func isNumber(u uint8) bool {
	return u >= '0' && u <= '9'
}