給定一個(gè)鏈表,在一定時(shí)間內(nèi)反轉(zhuǎn)這個(gè)鏈表的結(jié)點(diǎn),并返回修改后的鏈表。
如果結(jié)點(diǎn)數(shù)不是K的倍數(shù),那么剩余的結(jié)點(diǎn)就保持原樣。
你不應(yīng)該在結(jié)點(diǎn)上修改它的值,只有結(jié)點(diǎn)自身可以修改。
只允許使用常量空間。
例如
給定鏈表: 1->2->3->4->5
對(duì)于 k = 2,你應(yīng)該返回: 2->1->4->3->5
對(duì)于 k = 3,你應(yīng)該返回: 3->2->1->4->5
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
下面的代碼并不是我的,雖然我想的差不多,但就是這個(gè)差不多導(dǎo)致結(jié)果的差異性……
任重而道遠(yuǎn)??!
/**
* 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) {
auto node = head;
for(int i = 0; i < k; ++ i) {
if(!node) return head;
node = node->next;
}
auto new_head = reverse(head, node);
head->next = reverseKGroup(node, k);
return new_head;
}
ListNode* reverse(ListNode* start, ListNode* end) {
ListNode* head = end;
while(start != end) {
auto temp = start->next;
start->next = head;
head = start;
start = temp;
}
return head;
}
};