leetCode热题100-23、24题

23、链表反转

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
       ListNode prev = null;
       ListNode curr = head;
       while (curr != null) {
        ListNode next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
       }
       return prev;
    }
}

24、回文链表

给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        List<Integer> vals = new ArrayList<Integer>();

        // 将列表的值复制到数组中
        ListNode currentNode = head;
        while (currentNode != null) {
            vals.add(currentNode.val);
            currentNode = currentNode.next;
        }

        int front = 0,back = vals.size() - 1;
        while (front < back) {
            if (!vals.get(front).equals(vals.get(back))) {
                return false;
            }
            front++;
            back--;
        }
        return true;
    }
}