LeetCode 25: Reverse Nodes in k-Group

Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,

Given this linked list:

1->2->3->4->5

For k = 2, you should return:

2->1->4->3->5

For k = 3, you should return:

3->2->1->4->5

解题思路

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
if (k < 2) return head;

// 设置一个额外的表头指针
ListNode result(0), *tail = &result;

int count = 0;
ListNode *begin = head;
while (head) {
count++;

if (count == k) {
// 将 begin 和 head 之间的子链表反转，并链到 tail 结点后边，更新 tail 指针
ListNode *p = begin, *q = begin->next, *end = head->next;
while (q != end) {
ListNode *temp = q->next;
q->next = p;
p = q;
q = temp;
}

tail->next = p;
tail = begin;

count = 0;
head = begin = q;
}
else {
head = head->next;
}
}

tail->next = begin;
return result.next;
}
};