题目描述
给定一个链表的头结点head,请判断该链表是否为回文结构。例如: 1->2->1,返回true。1->2->2->1,返回true。15->6->15,返回true。1->2->3,返回false。
进阶:如果链表长度为N,时间复杂度为O(N),额外空间复杂度达到O(1)。
实现
public class IsPalindromeList {
public static class Node {
int val;
Node next;
Node(int data) {
this.val = data;
}
}
// 利用栈,需要n的额外空间
public static boolean isPalindrome1(Node head) {
Stack<Node> stack = new Stack<>();
Node cur = head;
while (cur != null) {
stack.push(cur);
cur = cur.next;
}
while (head != null) {
if (head.val != stack.pop().val) {
return false;
}
head = head.next;
}
return true;
}
// 快慢指针加栈,需要N/2额外空间
public static boolean isPalindrome2(Node head) {
if (head == null || head.next == null) return true;
Node slow = head.next; // 这样保证慢指针在后半部分的起点位置
Node fast = head;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
Stack<Node> stack = new Stack<>();
while (slow != null) {
stack.push(slow);
slow = slow.next;
}
while (!stack.isEmpty()) {
if (head.val != stack.pop().val) {
return false;
}
head = head.next;
}
return true;
}
// 空间复杂度O(1)
public static boolean isPalindrome3(Node head) {
if (head == null || head.next == null) return true;
// 找到中间节点
Node n1 = head;
Node n2 = head;
while (n2.next != null || n2.next.next != null) {
n1 = n1.next;
n2 = n2.next.next;
}
// 反转右边链表
n2 = n1.next;// n2 指向右边第一个节点
n1.next = null; // mid.next -> null
Node n3 = null;
while (n2 != null) {
n3 = n2.next;
n2.next = n1;
n1 = n2;
n2 = n3;
}
// 比较两个链表
n3 = n1; // save last node
n2 = head; // n2 -> left first node
boolean res = true;
while (n1 != null || n2 != null) {
if (n1.val != n2.val) {
res = false;
break;
}
n1 = n1.next;
n2 = n2.next;
}
// 还原链表
n1 = n3.next;
n3.next = null;
while (n1 != null) {
n2 = n1.next;
n1.next = n3;
n3 = n1;
n1 = n2;
}
return res;
}
public static void main(String[] args) {
Node head = null;
System.out.println(isPalindrome2(head));
head = new Node(1);
System.out.println(isPalindrome2(head));
head = new Node(1);
head.next = new Node(2);
System.out.println(isPalindrome2(head));
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(2);
head.next.next.next.next = new Node(1);
System.out.println(isPalindrome2(head));
}
}