三元运算符（Leetcode 439）

思路：
递归分解的思路很简单，但每次都要遍历整个string看?对应的:在哪里，因此可以先一边遍历一边使用栈记录问号?对应的分号:在哪里，这样就减少了每次遍历。
输入：“T?2:3”
输出：“2”
输入： “F?1:T?4:5”
输出： “4”
输入：″T?T?F:5:3″
输出： “F”

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String line = scanner.nextLine();

        System.out.println(calcExpress(line));
    }

    private static String calcExpress(String line) {
        int[] semicolonIndexes = new int[line.length()];
        Stack<Integer> stack = new Stack<>();

        for (int i = 0; i < line.length(); i++) {
            char ch = line.charAt(i);
            if (ch == '?') {
                stack.push(i);
            } else if (ch == ':') {
                int questionMarkIndex = stack.pop();
                semicolonIndexes[questionMarkIndex] = i;
            } else {
                continue;
            }
        }

        return calcExpress(line, semicolonIndexes, 0, line.length());
    }

    private static String calcExpress(String line, int[] semicolonIndexes, int start, int end) {
        if (end - start == 1) {
            return line.substring(start, end);
        }

        for (int i = start; i < end; i++) {
            char ch = line.charAt(i);

            if (ch == '?') {
                int semicolonIndex = semicolonIndexes[i];
                return line.charAt(i - 1) == 'T' ? calcExpress(line, semicolonIndexes, i + 1,
                    semicolonIndex) : calcExpress(line, semicolonIndexes, semicolonIndex + 1, end);
            }
        }

        return "impossible result";
    }