leetcode -. 206 inverted list

topic:

Reverse a singly linked list.
Example:
Input: 1-> 2-> 3-> 4- > 5-> NULL
Output: 5-> 4-> 3-> 2- > 1-> NULL

A thought:

Reversed pointer, three pointers, pointing to the front after sequentially reversed pointer, head and tail point to the original
special cases should be noted: [], [1]

  public ListNode reverseList(ListNode head) {
     ListNode pre = head;    //前一个指针
      ListNode nex;           //后一个指针
      if(head == null)        //处理输入空链表的情况
          return head;
      else
         nex = head.next;
      pre.next = null;        //将原来的头节点变为尾节点,将他的next置空
      while(nex != null){
          head = nex;
          nex = head.next;    //移动指针
          head.next = pre;    //更改结点指针的指向,将指向后面的指针变为指向前面
          pre = head;
      }
      return head;
  }

Refer leetcode Gangster Comments

  public static ListNode reverseListIterative(ListNode head) {
      ListNode prev = null; //前指针节点
      ListNode curr = head; //当前指针节点
      //每次循环，都将当前节点指向它前面的节点，然后当前节点和前节点后移
      while (curr != null) {
          ListNode nextTemp = curr.next; //临时节点，暂存当前节点的下一节点，用于后移
          curr.next = prev; //将当前节点指向它前面的节点
          prev = curr; //前指针后移
          curr = nextTemp; //当前指针后移
      }
      return prev;
  }
  转自: https://leetcode-cn.com/problems/reverse-linked-list/comments/

Thinking two:

Method to create a head pointer forward runs, the original, traversing the list, the original to the new node is inserted one by one in the list
of this new method is actually not a new node, except the original node, the above new pointer a row order, forward runs by a method, a new sequence is achieved in reverse order of the original

  public ListNode reverseList(ListNode head) {
     ListNode new_head = null;   //建立新指针
      while(head != null){
          ListNode tmp = head;    //建立临时结点保存head
          head = head.next;       //head后移
          tmp.next = new_head;    //将tmp前插到新的链表中
          new_head = tmp;
      }
  }
  return new_head;

Thinking three:

Recursive, recursive directly to the end of the list, pointing to the tail of p, then treated with a head, and finally return to the first layer p

public ListNode reverseList(ListNode head) {
 if(head==null || head.next==null) return head;  //处理[],[1]两种情况,将head指针定位在原来链表尾结点的前一个结点,然后进行处理
     ListNode p = reverseList(head.next);    //将p定位在原来链表的尾部,不需要对p进行其他操作
     head.next.next = head;  //  反转链表指针
     head.next = null;
     return p;
}