LinkedList - 206. Reverse Linked List

206. Reverse Linked List

Reverse a singly linked list.

Example:

**Input:** 1->2->3->4->5->NULL
**Output:** 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

思路：

翻转链表。主要有两种方法，迭代和递归。递归就是不断的把当前节点指向前一个链表的新头节点，递归出口是当前节点为nil。迭代的做法是两个指针往后滑动，保存好当前节点的下一个节点，避免断链。

代码：

golang:

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseListRecurese(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    
    return reverse(head, nil)
}

func reverse(head, newHead *ListNode) *ListNode {
    if head == nil {
        return newHead
    }
    
    tmp := head.Next
    head.Next = newHead
    return reverse(tmp, head)
}

func reverseList(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    
    var newHead *ListNode
    for head != nil {
        tmp := head.Next
        head.Next = newHead
        
        newHead = head
        head = tmp
    }
    
    return newHead
}

java：

/**

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

    private ListNode reverseListInt(ListNode head, ListNode newHead) {
        if (head == null)
            return newHead;
        ListNode next = head.next;
        head.next = newHead;
        return reverseListInt(next, head);
    }
}```