【Leetcode】Reverse Nodes in k-Group

问题

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

代码

class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
if (!head) {
return NULL;
}

ListNode *step = head;

ListNode *returnNode = NULL;
ListNode *lastNode = NULL;

while (step) {
ListNode *probe = step;
int i = k -1 ;
while (i > 0 && probe) {
probe = probe->next;
//step = step->next;
i--;
}

if (i != 0 || !probe) {
/* get all*/
break;
}

ListNode *front = step;
ListNode *end = probe->next;
ListNode *end1 = probe->next;
/* good */
if (!returnNode) {
returnNode = probe;
}
else{
lastNode->next = probe;
}
lastNode = step;

i = k;
while (i -- ) {
front = step;
step = step->next;
front->next = end;
end = front;
}

step = end1;
}
if(!returnNode ) return head;
return returnNode;

}
};

分析

首先用探针找到尾部，然后记录下一个节点和上一个节点，然后内部进行翻转即可。

总结

n/a