385. Mini Parser

class Solution {
    private static final NestedInteger DELIMITER = new NestedInteger(0);

    public NestedInteger deserialize(String s) {
        Stack<NestedInteger> stack = new Stack<>();

        int sign = 1;

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '-') {
                sign = -1;
            }

            if (Character.isDigit(c)) {
                int num = sign * (c - '0');
                while (i + 1 < s.length() && Character.isDigit(s.charAt(i + 1))) {
                    num = num * 10 + sign * (s.charAt(i + 1) - '0');
                    i++;
                }
                sign = 1;
                stack.push(new NestedInteger(num));
            } else if (c == '[') {
                NestedInteger ni = new NestedInteger();
                stack.push(ni);
                stack.push(DELIMITER);
            } else if (c == ']') {
                List<NestedInteger> niList = new ArrayList<>();
                while (stack.peek() != DELIMITER) {
                    niList.add(stack.pop());
                }
                stack.pop();
                NestedInteger root = stack.peek();
                for (int j = niList.size() - 1; j >= 0; j--) {
                    root.add(niList.get(j));
                }
            }
        }

        return stack.pop();
    }
}

该题题目阐述不清，并未说明 NestedInteger 不能同时装 num 与 NestedInteger，而是需要将每个 num 包装为 NestedInteger。

References

385. Mini Parser