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题目:
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.
思路:
找到m节点,从节点m到n依次反转指针,然后把翻转后的串连起来即可
/** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} head * @param {number} m * @param {number} n * @return {ListNode} */ var reverseBetween = function(head, m, n) { if(head==null){ return head; } var p=head,mpre=null; var tempHead=new ListNode(0); tempHead.next=head; mpre=tempHead; for(var i=1;i<m;i++){ mpre=p; p=p.next; } var pre=null,cur=null,next=null; var tempp=p; for(var i = 1; i <= n-m; i++){//反转m到n的指针 pre = p; cur = p.next; next = cur.next; cur.next = pre; p=cur; } mpre.next=p; tempp.next=next; head=tempHead.next; tempHead=null; return head; };
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原文地址:http://www.cnblogs.com/shytong/p/5143669.html