删除指定值的结点

Remove all elements from a linked list of integers that have value val.

Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5

思路与前面删除重复节点一致！

public class Solution {    public ListNode removeElements(ListNode head, int val) {        ListNode fakeHead = new ListNode(-1);        fakeHead.next = head;        ListNode curr = head, prev = fakeHead;        while (curr != null) {            if (curr.val == val) {                prev.next = curr.next;            } else {                prev = prev.next;            }            curr = curr.next;        }        return fakeHead.next;    }}

public class Solution {    public ListNode removeElements(ListNode head, int val) {        if (head == null) return null;        head.next = removeElements(head.next, val);        return head.val == val ? head.next : head;    }}

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 ...

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