湖北正规网站建设检修网站ui设计包括哪些原则

题目链接

• 反转链表：206. 反转链表 - 力扣（LeetCode）
• K个一组翻转链表：25. K 个一组翻转链表 - 力扣（LeetCode）
• 回文链表：234. 回文链表 - 力扣（LeetCode）

实现思路

代码实现

• 反转链表：

/*** 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* reverseList(ListNode* head) {ListNode* pre = nullptr;ListNode* cur = head;while(cur) {ListNode* next = cur -> next;cur -> next = pre;pre = cur;cur = next;}return pre;}
};

• K个一组翻转链表

/*** 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:vector<ListNode*> reverse(ListNode* head, ListNode* next) {ListNode* pre = next;ListNode* cur = head;while (cur != next) {ListNode* nextNode = cur -> next;cur -> next = pre;pre = cur;cur = nextNode;}return {pre, head}; // pre是反转之后的头节点,head是反转之后的尾节点}ListNode* reverseKGroup(ListNode* head, int k) {int cnt = 0;ListNode* cur = head;ListNode* preHead = new ListNode(-1);preHead -> next = head;ListNode* pre = preHead;while (cur) {cnt++;if (cnt % k == 0) {cnt = 0;vector<ListNode*> res = reverse(pre->next, cur->next);pre->next = res[0];cur = res[1];pre = cur;}cur = cur -> next;}return preHead -> next;}
};

• 回文链表

/*** 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* reverse(ListNode* head) {ListNode* pre = nullptr;ListNode* cur = head;while (cur) {ListNode* next = cur -> next;cur -> next = pre;pre = cur;cur = next;}return pre;}bool isPalindrome(ListNode* head) {if (head -> next == nullptr) return true;ListNode* preHead = new ListNode();preHead -> next = head;ListNode* slow = preHead;ListNode* fast = preHead;while (fast != nullptr && fast -> next != nullptr) {slow = slow -> next;fast = fast -> next -> next;}// 反转slow->next到tail的部分ListNode* head1 = slow -> next;head1 = reverse(head1);ListNode* cur = head;ListNode* cur1 = head1;while (cur1) {if (cur -> val != cur1 -> val) {return false;}cur = cur -> next;cur1 = cur1 -> next;}return true;}
};