一、反转链表 II
/**
* 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 || !head->next)
return head; ListNode* dummy = new ListNode();
dummy->next = head; ListNode* prev = dummy;
for (int i = ; i < m - ; i++)
prev = prev->next;
ListNode* start = prev->next;
ListNode* then = start->next; for (int i = ; i < n - m; i++)
{
start->next = then->next;
then->next = prev->next;
prev->next = then;
then = start->next;
}
return dummy->next;
}
};
二、回文链表
【题目】234. 回文链表
class Solution {
public:
bool isPalindrome(ListNode* head)
{
if (!head || !head->next) return true;
ListNode *slow = head, *fast = head;
while (fast->next && fast->next->next) //找到中点
{
slow = slow->next;
fast = fast->next->next;
}
ListNode *last = slow->next, *pre = head;
while (last->next) //反转
{
ListNode *tmp = last->next;
last->next = tmp->next;
tmp->next = slow->next;
slow->next = tmp;
}
while (slow->next) //比较
{
slow = slow->next;
if (pre->val != slow->val)
return false;
pre = pre->next;
}
return true;
}
};三、两两交换链表中的节点
【题目】24. 两两交换链表中的节点
C++ Soution 1:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head)
{
if (!head || !head->next) return head;
ListNode *t = head->next;
head->next = swapPairs(head->next->next);
t->next = head;
return t;
}
};
C++ Soution 2:
class Solution {
public:
ListNode* swapPairs(ListNode* head)
{
ListNode *dummy = new ListNode(-), *pre = dummy;
dummy->next = head;
while (pre->next && pre->next->next)
{
ListNode *t = pre->next->next;
pre->next->next = t->next;
t->next = pre->next;
pre->next = t;
pre = t->next;
}
return dummy->next;
}
};