linkedlist

链表问题包含：
删除节点，

插入排序，

旋转

Remove Linked List Elements

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param head, a ListNode    # @param val, an integer    # @return a ListNode    def removeElements(self, head, val):        node=ListNode(0)#先给一个DUMMY HEAD        node.next=head        cur=node#指针        while cur and cur.next:            if cur.next.val==val:#当指针下一个为VAL                cur.next=cur.next.next#给下一个            else:                cur=cur.next#循环        return node.next#返回后面节点

Java

Insertion Sort List

30% Accepted
Sort a linked list using insertion sort.
Example
Given 1->3->2->0->null, return 0->1->2->3->null.

/** * Definition for ListNode. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int val) { *         this.val = val; *         this.next = null; *     }//1,dummyhead->while loop->swap */ public class Solution {    public ListNode insertionSortList(ListNode head) {        ListNode dummy = new ListNode(0);//dummyhead        while (head != null) {//head is not null            ListNode node = dummy;//node is dummy            while (node.next != null && node.next.val < head.val) {                node = node.next; //当符合要求的时候，往下走            }            //不符合要求对调swap            //node.next->head.next            ListNode temp = head.next;//head.next为TEMP            head.next = node.next;//            node.next = head;            head = temp;        }        return dummy.next;    }}

双指针：
Remove Nth Node From End of List

两个指针，一个指针比另外一个快走N步，然后慢指针和快指针一起走。当快指针走到尽头时，正好，慢指针走到要删除的点。

class Solution:    # @return a ListNode    def removeNthFromEnd(self, head, n):        dummy = ListNode(0);        dummy.next = head;        fast = dummy;        for i in range(0,n):            if fast == None or n==0:                return head            if fast != None:                fast = fast.next        slow = dummy        while fast.next != None:            fast = fast.next            slow = slow.next        slow.next = slow.next.next        return dummy.next