Problem: 25. K 个一组翻转链表

题目描述

思路

1.创建虚拟头节点dummy并将其next指针指向head，创建指针pre、end均指向dummy；
2.编写反转单链表的函数reverse
3.当end -> next 不为空时：

3.1.每次k个一组将end指针后移；
3.2.判断当end指针为空时直接退出循环；
3.3.创建nextP指针指向end -> next,strat指针指向pre -> next;end的next指针置空；
3.4.从start指向的节点开始反转链表，并使得pre的next指针指向其；
3.5.start的next指针指向nextP，使得原始链表能连接起来
3.6.使pre指针指向start指针，end指针指向pre指针

复杂度

时间复杂度:

$O(n)$；其中$n$为链表的长度

空间复杂度:

$O(n)$

Code

/*** 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:/*** A set of K reversed single-linked lists* * @param head The head node of the list to be reversed* @param k Number of nodes to be reversed* @return ListNode**/ListNode* reverseKGroup(ListNode* head, int k) {//Create a virtual node ListNode* dummy = new ListNode(INT_MIN);ListNode* pre = dummy;ListNode* end = dummy;dummy -> next = head;while (end -> next != nullptr) {//Each k is in a groupfor (int i = 0; i < k && end != nullptr; ++i) {//Pointer end moves backend = end -> next;}if (end == nullptr) {break;}ListNode* start = pre -> next;ListNode* nextP = end -> next;//disconnectend -> next = nullptr;//reverse k nodes and return the new headnode which //is the last node of k nodespre -> next = reverse(start);//connect next group k nodes by next pointerstart -> next = nextP;pre = start;end = pre;}return dummy -> next;}/*** Reverse the single linked list* * @param head The node of the list to be reversed* @return ListNode**/ListNode* reverse(ListNode* head) {ListNode* pre = nullptr;ListNode* cur = head;while (cur != nullptr) {ListNode* nextP = cur -> next;cur -> next = pre;pre = cur;cur = nextP;}return pre;}
};