leetcode025：Reverse Nodes in k-Group

问题描述

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

问题分析

以k个结点为一组，逆转组内结点后连接起来，是leetcode024的升级版本（k=2）。解题思路比较清晰，先查看是否有足够的结点构成一组，有则逆转，不够则直接连接起来。难点是链表操作的先后复制要清晰，否则容易混乱。下面以链表为12345678，k=3为例说明链表复制的过程（链表逆转逻辑比较简单，这里不提）。

ans是最终结果的表头指针；
ans_p是负责组间连接工作的结点指针；
p是负责检查是否还有k个结点需要逆转的结点指针；
p1用来记录逆转前的组的表头指针；
q为逆转过程中的表头指针。

代码

<pre name="code" class="cpp">//运行时间：38ms
class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
if (!head) return NULL;
if (k<=1) return head;
ListNode *ans = NULL;
ListNode *ans_p = head;
ListNode *p = head;//前进！
ListNode *p1;
ListNode *q = head;
ListNode *x;
ListNode *y;
int count ;
while (1){
count = 0;
p1 = p;
while (p&&count < k){
count++;
p = p->next;
}
if (count < k) {
if (!ans)	ans = p1;
else{ ans_p->next = p1; }
break;
}
else{
x = y = q->next;
count = 1;
while (count < k){
x = x->next;
y->next = q;
q = y;
y = x;
count++;
}
if (!ans) { ans = q; ans_p = p1; }
else { ans_p->next = q; ans_p = p1; }
q = p;
}
if (!p) break;
}
return ans;
}
};