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Reverse Linked List II

时间:2014-12-03 22:44:08      阅读:134      评论:0      收藏:0      [点我收藏+]

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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.

/**
 * 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) {
        int i;
        ListNode *p, *p1, *p2,*p3,*p4,*tmp;
        p = p1 = p2 = p3 = p4 = tmp = NULL;
        for (i=1, p=head; p; i++, p=p->next){
            if(i==m-1) p1 = p;
            if(i==m) p2 = p;
            if(i==n) p3 = p;
            if(i==n+1) p4 = p;
        }

        while(p2!=p3){
            tmp = p2->next;
            p2->next = p4;
            p4 = p2;
            p2 = tmp;
        }
        p2->next = p4;
        p4 = p2;

        if(p1){
            p1->next = p4;
        } else {
            head = p4;
        }

        return head;
    }
};

 

Reverse Linked List II

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原文地址:http://www.cnblogs.com/code-swan/p/4141190.html

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