首页 > 其他 > 详细

[LeetCode] Reverse Linked List II

时间:2014-08-13 00:50:34      阅读:325      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->NULL, m = 2 and n = 4,

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

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        if(m==n)
            return head;
        int num = 1;
        ListNode *h = head;
        ListNode *Pm = head,*Pm_pre=NULL;
        while(num != m){
            Pm_pre = Pm;
            Pm = Pm->next;
            num++;
        }//end while
        ListNode *p1 = Pm,*p2 = p1->next,*p3 = NULL;
        while(num!=n){
           p3 = p2->next;
           p2->next = p1;
           p1 = p2;
           p2 = p3;
           num++;
        }//end while
        if(Pm_pre!=NULL){
          Pm_pre->next = p1;
          Pm->next   = p2;
        }else{
          h = p1;
          Pm->next = p2;
        }
        return h;
    }
};

 

[LeetCode] Reverse Linked List II,布布扣,bubuko.com

[LeetCode] Reverse Linked List II

原文:http://www.cnblogs.com/Xylophone/p/3908604.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!