Reverse Linked List II -- leetcode

标签：leetcode   reverse   倒序   链表   

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.

/**
 * 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) {
        ListNode fake(0);
        fake.next = head;
        ListNode *h = &fake;
        n -= m;
        while (--m && h)
            h = h->next;
            
        if (!h || !h->next) return fake.next;
        
        ListNode *last = h->next;
        ListNode *p = last->next;
        while (p && n--) {
            last->next = p->next;
            p->next = h->next;
            h->next = p;
            p = last->next;
        }
        
        return fake.next;
    }
};

Reverse Linked List II -- leetcode

标签：leetcode   reverse   倒序   链表   

原文地址：http://blog.csdn.net/elton_xiao/article/details/45335635