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;
}
}
};