1.单向链表的原地反转

public class A {    public A next;    private int n;    public A(int n) {        this.n = n;    }    public static A reverse(A a){        A current = a.next;        a.next = null;        A last = a;        while(true){            if (current == null) {                break;            }else{                A c = current.next;                current.next = last;                last = current;                current = c;            }        }        return last;    }    public static void main(String[] args) {        A a1 = new A(1);        A a2 = new A(2);        a1.next = a2;        A a3 = new A(3);        a2.next = a3;        A last = reverse(a1);        System.out.println(last.n);        System.out.println(last.next.n);        System.out.println(last.next.next.n);    }}

2.消除字符串中成对的括号，必须成对匹配，否则打印error

public class B{    public static void removeBrackets(char[] arr){        int leftBracket = 0;        StringBuilder sb = new StringBuilder();        for (char ch : arr) {            if (ch == '(') {                leftBracket++;            }else if(ch == ')'){                if (leftBracket == 0) {                    System.out.println("error");                    return;                }else{                    leftBracket --;                }            }else{                sb.append(ch);            }        }        if ( leftBracket != 0 ) {            System.out.println("error");            return;        }        System.out.println(sb.toString());    }    public static void main(String[] args) {        String s = "1(23456(789)abc)";        removeBrackets(s.toCharArray());    }}

3.将长度为a的单向链表链接到长度为b的单向链表，算法复杂度是 A. O(blogb) B. O(b) C.O(a*b) D. O(logb)