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

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

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return?1->4->3->2->5->NULL.

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

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tag: two pointer

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/*** 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) {if(head == null || head.next == null|| m >= n || m <= 0 || n <= 0 ) {return head;}ListNode dummy = new ListNode(0);dummy.next = head;head = dummy;for(int i = 1; i < m; i++) {if(head == null) {return null;}head = head.next;}ListNode preM = head;ListNode mNode = head.next;ListNode nNode = mNode;ListNode nextN = mNode.next;for(int j = m; j < n; j++) {if(nextN == null) {return null;}ListNode tmp = nextN.next;nextN.next = nNode;nNode = nextN;nextN = tmp;}preM.next = nNode;mNode.next = nextN;return dummy.next;} }

　　

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轉載于:https://www.cnblogs.com/superzhaochao/p/6583121.html

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