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)