leetcode题解-328. Odd Even Linked List && 19. Remove Nth Node From End of List

328，题目

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 ...

本体是要将链表的第奇偶的元素分开，讲奇数元素放在前面，偶数元素放在后面。其实思路十分简单，使用两个指针遍历奇偶元素即可，然后修改相应元素的next指针指向位置。代码如下所示：

public ListNode oddEvenList(ListNode head) {
if(head == null || head.next == null)
return head;
ListNode ji=head, sec=head.next, ou = sec;
while(ji.next != null && ou.next != null){
ji.next = ou.next;
ou.next = ji.next.next;
ji = ji.next;
ou = ou.next;
}
if(ou != null)
ou.next = null;
ji.next = sec;
return head;
}

19， 题目：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.

本题是要删除倒数第n个元素，其实是两个指针的路程差问题，思路是使用fast和slow两个指针，fast先走n步，slow指针再走，遮阳挡fast走到最后的时候，slow也就到了倒数第n+1个，然后将该元素删除就行了。代码如下所示：

public ListNode removeNthFromEnd(ListNode head, int n) {

ListNode start = new ListNode(0);
ListNode slow = start, fast = start;
slow.next = head;

//Move fast in front so that the gap between slow and fast becomes n
for(int i=1; i<=n+1; i++)   {
fast = fast.next;
}
//Move fast to the end, maintaining the gap
while(fast != null) {
slow = slow.next;
fast = fast.next;
}
//Skip the desired node
slow.next = slow.next.next;
return start.next;
}