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Reverse Linked List II

时间:2014-02-16 20:03:59      阅读:373      评论:0      收藏:0      [点我收藏+]

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->NULLm = 2 and n = 4,

return 1->4->3->2->5->NULL.

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

 ref: http://www.cnblogs.com/feiling/p/3263501.html

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(head == null || m == n)
            return head;
        
        ListNode fake = new ListNode(Integer.MIN_VALUE);
        fake.next = head;
        head = fake;
        
        int count = 0;
        ListNode pre = null, p = head, p1 = null, p2 = null, p3 =null;
        while(count <= n){
            if(count == m-1){
                p1 = p;
            }else if(count == m){
                p2 = p;
            }else if(count == n){
                p3 = p;
            }
            
            ListNode tmp = p.next;
            
            if(count >= m+1 && count <= n){
                p.next = pre;
            }
            pre = p;
            p = tmp;
            count++;
        }
        p1.next = p3;
        p2.next = p;
        
        return fake.next;
    }
}
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Reverse Linked List II

原文:http://www.cnblogs.com/RazerLu/p/3551085.html

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