LeetCode 328: Odd Even Linked List

328. Odd Even Linked List

Difficulty: Medium

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example:

Given 1->2->3->4->5->NULL,

return 1->3->5->2->4->NULL.

Note:

The relative order inside both the even and odd groups should remain as it was in the input.

The first node is considered odd, the second node even and so on …

思路

首先，最重要的是看清题目。这道题只看例子的话，很容易会以为是对节点的值进行排序，使奇数值在偶数值前。其实，是将排在奇数位的节点放在偶数位的节点前。

于是，除指向当前遍历的节点的指针node外，还需要两个指针odd和even来分别指向已遍历节点中最后一个排奇数位和排偶数位的节点。

改变节点排列时，需要注意顺序，node先删除再插入（node为even的下一节点）。

1、even的下一节点为node的下一节点；

2、node的下一节点为odd的下一节点；

3、odd的下一节点为node。

代码

[C++]

class Solution {
public:
ListNode* oddEvenList(ListNode* head) {
if(head == NULL)
return head;
ListNode *odd = head;
ListNode *even = odd->next;
ListNode *node = NULL;
while(even && even->next) {
node = even->next;
even->next = node->next;
node->next = odd->next;
odd->next = node;
odd = odd->next;
even = even->next;
}
return head;
}
};