92. 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.

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思路：

    设置一个辅助指针指向链表的头结点。首先定位到需要进行翻转的链表的开头，pre对应中间这段链表头结点的前一个结点。然后就地对中间这段链表进行翻转处理，保持pre指向翻转过后的链表的头部。这样最后辅助指针的next结点就是处理过后的链表的头部。

    请注意翻转时指针的连接处理。

           begin->next=then->next;

           then->next=pre->next;

           pre->next=then;

           then=begin->next;

/**
 * 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 *dumpy =new ListNode (-1);
        dumpy ->next= head;
        int i;
        ListNode *pre =dumpy;
        for(i=0;i<m-1;i++)
            pre=pre->next;
        ListNode *begin = pre->next;
        ListNode *then = begin->next;
        for(i=0;i<n-m;i++){
           begin->next=then->next;
           then->next=pre->next;
           pre->next=then;
           then=begin->next;
        }
        return dumpy->next;
    }
};

参考代码：java

public ListNode reverseBetween(ListNode head, int m, int n) {
    if(head == null) return null;
    ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list
    dummy.next = head;
    ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing
    for(int i = 0; i<m-1; i++) pre = pre.next;

    ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed
    ListNode then = start.next; // a pointer to a node that will be reversed

    // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3
    // dummy-> 1 -> 2 -> 3 -> 4 -> 5

    for(int i=0; i<n-m; i++)
    {
        start.next = then.next;
        then.next = pre.next;
        pre.next = then;
        then = start.next;
    }

    // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
    // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)

    return dummy.next;

}