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.

用一个辅助节点helper找到m的位置,然后依次交换位置,代码如下:
/**
 * 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) {
        ListNode helper = new ListNode(0);
        helper.next = head;
        head = helper;
        int k = n - m;
        while(m > 1) {
            head = head.next;
            m --;
        }
        ListNode reve = head.next;
        while(k > 0) {
            ListNode cur = reve.next;
            reve.next = cur.next;
            cur.next = head.next;
            head.next = cur;
            k --;
        }
        return helper.next;
    }
}

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转载自kickcode.iteye.com/blog/2275984
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