[leetcode] 328. Odd Even Linked List

Description

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 1:

Input:

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

Output:

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

Example 2:

Input:

 2->1->3->5->6->4->7->NULL

Output:

2->3->6->7->1->5->4->NULL

Note:

1. The relative order inside both the even and odd groups should remain as it was in the input.
2. The first node is considered odd, the second node even and so on …

分析

题目的意思是：把链表中需要为奇数的节点找出来组成链表，要求O(1)空间复杂度，O(nodes)时间复杂度。

• 思路很直接，就是边遍历边构成奇索引链表。这里要注意断开的顺序，防止空指针的情况，先让节点从原来的链表中独立出来，原来的链表要保持不断，然后再插入到奇数节点链表中。模拟一下就可以了。

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(!head||!head->next){
            return head;
        }
        ListNode* odd=head;
        ListNode* even=head->next;
        ListNode* even_head=even;
        while(even&&even->next){
            odd->next=even->next;
            odd=odd->next;
            even->next=odd->next;
            even=even->next;
        }
        odd->next=even_head;
        return head;
    }
};

参考文献

​​[LeetCode] Odd Even Linked List 奇偶链表​​