LeetCode-206&92.Reverse Linked List

206 https://leetcode.com/problems/reverse-linked-list/

Reverse a singly linked list.

/** * Definition for singly-linked list. * public class ListNode { *     public int val; *     public ListNode next; *     public ListNode(int x) { val = x; } * } */public class Solution{    public ListNode ReverseList(ListNode head)     {        if (head == null || head.next == null)            return head;        ListNode pre=null, next;        while (head != null)        {            next = head.next;            head.next = pre;            pre = head;            head = next;        }        return pre;    }}

递归

public ListNode ReverseList(ListNode head)        {            if (head == null || head.next == null)                return head;            ListNode next = head.next;            head.next = null;            ListNode node = ReverseList(next);            next.next = head;            return node;        }

92 https://leetcode.com/problems/reverse-linked-list-ii/

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */

使用栈：（低效）

public ListNode ReverseBetween(ListNode head, int m, int n)     {        if(head.next==null||m==n)            return head;                   ListNode pre = new ListNode(0);            pre.next = head;            ListNode curNode = pre;            Stack<ListNode> stack = new Stack<ListNode>();            for (int i = 0; i < m; i++)            {                curNode = curNode.next;            }            for (int i = m; i <= n; i++)            {                stack.Push(curNode);                curNode = curNode.next;            }            ListNode leftNode = curNode;            curNode = pre;            for (int i = 0; i < m - 1; i++)            {                curNode = curNode.next;            }            while (stack.Count!=0)            {                curNode.next = stack.Pop();                curNode = curNode.next;            }            curNode.next = leftNode;            return pre.next;    }

不用栈

public ListNode ReverseBetween(ListNode head, int m, int n)     {       if (head.next == null || m == n)                return head;            ListNode pre = new ListNode(0);            pre.next = head;            ListNode curNode = pre;            for (int i = 0; i < m - 1; i++)            {                curNode = curNode.next;            }            ListNode first = curNode.next;            ListNode next = first.next;            for (int i = m; i < n; i++)            {                first.next = next.next;                next.next = curNode.next;                curNode.next = next;                next = first.next;            }            return pre.next;    }