Leetcode 86. Partition List

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

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

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

Analysis:

1. Build both empty lists
2. val < x add it to the tail of first list
Val >= x add it to the tail of second list
3. Concatenate the tail of List1 to the head of List2
4. The tail of list2 List.next = null

Java Code:

20160601

/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution { PublicListNode partition (ListNode head,intx) {//1. Build both empty lists//2. val < x add it to the tail of first list//Val >= x add it to the tail of second list//3. Concatenate the tail of List1 to the head of List2//4. The tail of list2 List.next = null//Base Case        if(Head = =NULL|| Head.next = =NULL) {            returnHead; } ListNode cur=Head; ListNode dummy1=NewListNode (0); ListNode Head1=dummy1; ListNode Dummy2=NewListNode (0); ListNode head2=Dummy2;  while(cur! =NULL) {            if(Cur.val <x) {Dummy1.next=cur; Dummy1=Dummy1.next; } Else{Dummy2.next=cur; Dummy2=Dummy2.next; } cur=Cur.next; } Dummy2.next=NULL; Dummy1.next=Head2.next; returnHead1.next; }}

Leetcode 86. Partition List