LeetCode-92-Reverse Linked List II

算法描述:

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

解题思路:链表题,首先要画图。四个指针,头指针,前指针,临时指针。

    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* prev = dummy;
        ListNode* cur = dummy;
        for(int i=1; i < m; i++) 
            prev = prev->next;
        cur = prev->next;
        for(int i =m; i<n; i++){
            ListNode* temp = cur->next;
            cur->next = temp->next;
            temp->next =prev->next;
            prev->next = temp;
            
         }
        return dummy->next;
    }

 

posted on 2019-02-02 12:59  无名路人甲  阅读(72)  评论(0编辑  收藏  举报