在O(n) 时间复杂度，O(1)空间复杂度内反转单链表

在LeetCode中看到判断回文的程序：https://leetcode.com/problems/palindrome-linked-list/

里面用单链表来存储数据，先反转前半部分的单链表，然后分别从 表头 和 中间链表位置处 开始比较元素。

反转单链表的代码如下：

 1      private ListNode reverseList(ListNode head, int length){
 2          if(head == null)
 3              return null;
 4          ListNode currentNode, preNode;
 5          currentNode = preNode = head;
 6          ListNode nextNode = head.next;
 7          head.next = null;
 8          for(int i = 1; i < length; i++){
 9              if(nextNode != null){
10                  currentNode = nextNode;
11                  nextNode = nextNode.next;
12                  currentNode.next = preNode;
13                  preNode = currentNode;
14              }
15          }
16         return currentNode;
17      }

完整代码如下：

public class ReverseList {

    public static void main(String[] args) {
        ListNode head = new ListNode(0);
        ListNode node1 = new ListNode(1);
        head.next = node1;
        ListNode node2 = new ListNode(2);
        node1.next = node2;
        ListNode node3 = new ListNode(3);
        node2.next = node3;

        print(head);
        System.out.println();
        ListNode newHead = reverseList(head);
        print(newHead);
    }

    public static ListNode reverseList(ListNode head) {
        if (head == null) {
            return null;
        }

        ListNode currentNode,nextNode,preNode;
        currentNode = preNode = head;
        nextNode = head.next;

        head.next = null;
        while (nextNode != null) {
            currentNode = nextNode;
            nextNode = nextNode.next;
            //reverse
            currentNode.next = preNode;
            preNode = currentNode;
        }

        return currentNode;
    }

    private static class ListNode{
        private int val;
        private ListNode next;

        public ListNode(int val) {
            this.val = val;
        }
    }

    private static void print(ListNode head) {
        while (head != null) {
            System.out.print(head.val + " ");
            head = head.next;
        }
    }
}