【LEETCODE】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 …

这是一道奇偶链表题。可能因为很久都没有用过指针，链表这些，debug 了很久….

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

指针要注意的就是更新odd 和 even 的时候一定要想清楚它的具体位置。先把odd 的next 更新，再更新odd，再更新even-> next ，最后更新even。