Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

Solution: 建立两个链表，一个存值比x小的，一个存值比x大的，然后将两个链表连起来。

要将值大的那个表的最后一个数的next = null。

在建链表的时候我采用了头指针的方式。（-1那个指针）

 1 /**

 2  * Definition for singly-linked list.

 3  * public class ListNode {

 4  *     int val;

 5  *     ListNode next;

 6  *     ListNode(int x) {

 7  *         val = x;

 8  *         next = null;

 9  *     }

10  * }

11  */

12 public class Solution {

13     public ListNode partition(ListNode head, int x) {

14         // Start typing your Java solution below

15         // DO NOT write main() function

16         if(head == null) return head;

17         ListNode small = new ListNode(-1);

18         ListNode large = new ListNode(-1);

19         ListNode scur = small;

20         ListNode lcur = large;

21         while(head != null){

22             if(head.val < x){

23                 scur.next = head;

24                 scur = head;

25             }else{

26                 lcur.next = head;

27                 lcur = head;

28             }

29             head = head.next;

30         }

31         scur.next = large.next;

32         lcur.next = null;

33         return small.next;

34     }

35 }

 第二遍： 就在原来链表上修改。 将大于target的数字放到原来list的tail处。

 1 public class Solution {

 2     public ListNode partition(ListNode head, int x) {

 3         // IMPORTANT: Please reset any member data you declared, as

 4         // the same Solution instance will be reused for each test case.

 5         ListNode header = new ListNode(-1);

 6         header.next = head;

 7         int len = 0;

 8         ListNode cur = header, tail = null;

 9         while(cur.next != null){

10             cur = cur.next;

11             len ++;

12         }

13         tail = cur;

14         cur = header;

15         int i = 0;

16         while(i < len){

17             if(cur.next.val >= x){

18                 tail.next = cur.next;

19                 cur.next = cur.next.next;

20                 tail = tail.next;

21                 tail.next = null;

22             }else{

23                 cur = cur.next;

24             }

25             i ++;

26         }

27         return header.next;

28     }

29 }