(LeetCode) 328. Odd Even Linked List

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本文介绍了一种在不改变节点值的前提下，将单链表中的奇数位置节点和偶数位置节点进行分组的方法。该算法可在O(1)的空间复杂度和O(n)的时间复杂度下完成操作。

328. Odd Even Linked List

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 …

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==NULL) return head;
        ListNode* even_head = head->next;
        if (even_head==NULL) return head;

        ListNode* odd = head, *even = even_head;
        while(even!=NULL&&even->next!=NULL){
            odd->next = odd->next->next;
            even->next = even->next->next;

            odd = odd->next;
            even = even->next;
        }
        odd->next = even_head;
        return head;

    }
};