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  • 92. Reverse Linked List II 翻转链表II

    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

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode() : val(0), next(nullptr) {}
     *     ListNode(int x) : val(x), next(nullptr) {}
     *     ListNode(int x, ListNode *next) : val(x), next(next) {}
     * };
     */
    class Solution {
    public:
        ListNode* reverseBetween(ListNode* head, int m, int n) {
            ListNode* dummy = new ListNode(-1);
            dummy->next = head;
            ListNode* prem = dummy;
            //先找到第m个以及前一个
            for(int i=0;i<m-1;i++){
                prem = prem->next;
            }
            ListNode* cur = prem->next;
            //找到第n个以及后一个.
            ListNode* pren = dummy;
            for(int i=0;i<n;i++){
                pren = pren->next;
            }
            //翻转
            prem->next = reverse(cur,pren->next);
            //返回
            return dummy->next;
        }
        
        //@last 翻转链表的最后一个节点的后一个节点
        ListNode* reverse(ListNode* head,ListNode* last){
            ListNode *pre = last,*cur = head;
            while(cur != last){
                ListNode* Next = cur->next;
                cur->next = pre;
                pre = cur;
                cur = Next;
            }
            return pre;
        }
        
    };
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  • 原文地址:https://www.cnblogs.com/wsw-seu/p/13660622.html
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