題目
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.
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
分析
考察鏈表操作，思路并不復(fù)雜，就是非常繁瑣章钾，細(xì)心點辽社。
實現(xiàn)
/**
* 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) {
ListNode dummy(-1);
dummy.next = head;
ListNode *prev=&dummy, *cur, *tmp, *next, *end=head;
bool stop = false;
while(end!=NULL){
for(int i=0; i<k; i++){
if(end==NULL){
stop = true;
break;
}
end = end->next;
}
if(stop) break;
for(cur=prev->next, next=end;
cur->next!=end;
next = cur, cur = tmp, i++){
tmp = cur->next;
cur->next = next;
}
cur->next = next;
tmp = prev->next;
prev->next = cur;
prev = tmp;
end = prev->next;
}
return dummy.next;
}
};
思考
這種題要多練，熟練就好了福荸。
