25. Reverse Nodes in k-Group***
https://leetcode.com/problems/reverse-nodes-in-k-group/
题目描述
Given a linked list, reverse the nodes of a linked list k
at a time and return its modified list.
k
is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k
then left-out nodes in the end should remain as it is.
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
Note:
- Only constant extra memory is allowed.
- You may not alter the values in the list’s nodes, only nodes itself may be changed.
C++ 实现 1
迭代的方法. reverse
的方法不多说, 将 [head, end)
范围内的节点进行翻转. 再来看迭代的过程, 当写好了 reverse
函数, 之后就是确认各个大小为 k
的 Group, 用 n
来计数在 [head, tail]
范围内的节点的个数. (这段代码中 if (n == k) break;
放在哪里需要斟酌, 要想清楚). 当 while 结束时, 需要判断 n
是否等于 k
, 以便确认 [head, tail]
这个 Group 的节点个数是否达到 k
. 如果是的话, 那么就需要翻转 [head, tail->next)
范围内的节点. 否则就不用翻转. 之后更新 p
以及 head
也非常关键; 如果这个 Group 翻转了, 此时 head
将指向这个 Group 的最后一个节点, 因此, 此时 p
和 head
应该分别指向 head
以及 head->next
.
class Solution {
private:
ListNode* reverse(ListNode *head, ListNode *end) {
auto prev = end;
while (head != end) {
auto tmp = head->next;
head->next = prev;
prev = head;
head = tmp;
}
return prev;
}
public:
ListNode* reverseKGroup(ListNode* head, int k) {
ListNode *dummy = new ListNode(0);
auto p = dummy;
while (head) {
auto tail = head;
int n = 1;
// 统计在 [head, tail] 范围内节点的个数
while (tail->next) {
if (n == k) break;
++ n;
tail = tail->next;
}
if (n == k) p->next = reverse(head, tail->next);
else p->next = head;
p = head;
head = head->next;
}
return dummy->next;
}
};
C++ 实现 2
使用递归的方法. reverse
方法和 C++ 实现 1
相同. 下面代码中, 翻转 [head, tail)
内的节点.
class Solution {
private:
ListNode* reverse(ListNode *head, ListNode *end) {
auto prev = end;
while (head != end) {
auto tmp = head->next;
head->next = prev;
prev = head;
head = tmp;
}
return prev;
}
public:
ListNode* reverseKGroup(ListNode* head, int k) {
if (!head || !head->next) return head;
auto tail = head;
for (int i = 0; i < k; ++ i) {
if (!tail) return head; // 如果 Group 大小不够 k, 那么直接退出.
tail = tail->next;
}
auto newhead = reverse(head, tail); // 翻转后, head 为当前 Group 的最后一个节点
head->next = reverseKGroup(tail, k);
return newhead;
}
};
C++ 实现 3
如果忽视题目中关于额外空间的限制, 直接用 Stack …
class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
if (!head)
return nullptr;
stack<ListNode*> Stack;
auto end = head;
for (int i = 0; i < k; ++i) {
if (!end)
return head;
Stack.push(end);
end = end->next;
}
auto post = end;
ListNode *dummy = new ListNode(0);
auto path = dummy;
while (!Stack.empty()) {
path->next = Stack.top();
Stack.pop();
path = path->next;
}
path->next = reverseKGroup(post, k);
ListNode *res = dummy->next;
delete dummy;
return res;
}
};