Reverse Linked List I&&II——数据结构课上的一道题(经典必做题)

Reverse Linked List I

Question Solution

Reverse a singly linked list.

Reverse Linked List I

设置三个指针即可，非常简单：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head==NULL || head->next == NULL){
            return head;
        }
        ListNode* firstNode = head;
        ListNode* preCurNode = head;
        ListNode* curNode = head->next;//maybe null
        while(curNode){
            preCurNode->next = curNode->next;
            curNode->next = firstNode;
            firstNode=curNode;
            curNode =preCurNode->next;  

        }
        return firstNode;
    }
};

　　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.

/**
 * 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) {
        if(head==NULL||head->next==NULL||m==n)
            return head;
        ListNode* firstNode = head;
        ListNode* preCurNode = head;
        ListNode* curNode = head->next;//maybe null
        ListNode* lastNode= head;
        int flag=;
        int m_flag=flag;
        if(m==)
        {
            while(flag<n)
            {
                preCurNode->next = curNode->next;
                curNode->next = firstNode;
                firstNode=curNode;
                curNode =preCurNode->next;
                flag++;
            }
           return firstNode;
        }
        else
        {
            while(flag<n)
            {
              if(flag<m)
              {
                  lastNode=firstNode;
                 firstNode=firstNode->next;
                 preCurNode =preCurNode->next;
                 curNode =curNode->next;
                 flag++;
              }
              else
              {
                preCurNode->next = curNode->next;
                curNode->next = firstNode;
                firstNode=curNode;
                curNode =preCurNode->next;
                lastNode->next=firstNode;
                flag++;
               }

             }
         return head;
        }
    }
};