[LeetCode] Odd Even Linked List

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 ...

Time Complexity
O(N)
Space Complexity
O(N)

思路

Think about these test cases
Corner case : empty linked list, linked list with only one node, linked list with only two nodes, a linked list with odd number of nodes, a linked list with even number of nodes

Maintain two pointer 'odd' and 'even' for current nodes at odd position and even position

We'll also keep the first node of even linked list so that we can link the end of odd node together.

代码

public ListNode oddEvenList(ListNode head) {
    //corner case
    if(head == null || head.next == null) return head;
    ListNode odd = head;
    ListNode even = head.next;
    ListNode evenHead = head.next;
    while(even != null && even.next != null){
        odd.next = even.next;
        even.next = even.next.next;
        odd = odd.next;
        even = even.next;
    }
    odd.next = evenHead;
    return head;
}