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=1;
        int m_flag=flag;
        if(m==1)
        {
            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;     
        }
    }
};