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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
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思路：
/*
k=3
第一次反转，初始状态：
-1 -> 1 -> 2 -> 3 -> 4 -> 5
| | |
pre cur tmp
第1次操作后：
-1 -> 2 -> 1 -> 3 -> 4 -> 5
| | |
pre cur tmp
第2次操作后：
-1 -> 3 -> 2 -> 1 -> 4 -> 5
| | |
pre cur tmp
第二次反转，初始状态：
-1 -> 3 -> 2 -> 1 -> 4 -> 5
| | |
pre cur tmp
...
*/code:
/**
* 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(!head || k<=1) return head;
ListNode dummy(0);
dummy.next = head;
ListNode *pre = &dummy;
ListNode *cur = head;
ListNode *tmp;
int len = 0;
while(cur){
len++;
cur = cur->next;
}
while(len>=k){
cur = pre->next;
tmp = cur->next;
for(int i=1;i<k;i++){
cur->next = tmp->next;
tmp->next = pre->next;
pre->next = tmp;
tmp = cur->next;
}
pre = cur;
len -= k;
}
return dummy.next;
}
};LeetCode:Reverse Nodes in k-Group
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原文地址：http://blog.csdn.net/itismelzp/article/details/51167494
