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.

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解题思路:

反转整个链表的变种，指定了起点和终点。由于m=1时会变动头节点，所以加入一个dummy头节点

 

1. 找到原链表中第m-1个节点start：反转后的部分将接回改节点后。

从dummy开始移动m-1步

 

D->1->2->3->4->5->NULL

       |

      st

 

2. 将从p = start->next开始，长度为L = n-m+1的部分链表反转。

            __________

            |                  |

            |                 V

D->1->2<-3<-4    5->NULL             

       |     |           | 

      st    p          h0         

 

3. 最后接回

 

            __________

            |                  |

            |                 V

D->1   2<-3<-4    5->NULL             

       |________|                

 

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Java code:

20160601

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseBetween(ListNode head, int m, int n) {        //iterative        //base case        if (head == null || head.next == null || m < 1 || m >= n) {            return head;        }        ListNode dummy = new ListNode(0);        dummy.next = head;        ListNode pre = dummy;        ListNode cur = head;        //move m-1 to the node before reverse start         int count = 1;        while (count < m) {            pre = cur;            cur = cur.next;            count++;        }        ListNode beforeReverse = pre;        ListNode startReverse = cur;                while (count <= n) {            ListNode next = cur.next;            cur.next = pre;            pre = cur;            cur = next;            count++;        }        beforeReverse.next = pre;        startReverse.next = cur;        return dummy.next;    }}

 

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Reference:

1. http://bangbingsyb.blogspot.com/2014/11/leetcode-reverse-linked-list-ii.html