[Leetcode]86. Partition List@java Problem Solving report __java

https://leetcode.com/problems/partition-list/description/

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

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

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

Given a single linked list and an X, put less than X in the list to the front, greater than or equal to X to the back, the original relative position of each element is unchanged.

Train of thought: new two nodes preHead1 and preHead2, respectively, point to two linked list of the head node.

A node with a node value of less than X is linked to the list 1, and nodes with a node value equal to X are linked to the list 2.

Finally, two linked lists can be connected

Package go.jacob.day817;

/**
 * 86. Partition List
 * 
 * @author Jacob
 *
/public class Demo3 {public
	listnode Partition (listnode Head, int x) {
		//prehead1 and preHead2 are the forward nodes of two linked header nodes (one linked list has a node value less than x, another greater than equal x)
		ListNode preHead1 = new ListNode (0), PreHead2 = new ListNode (0);
		ListNode cur1 = preHead1, cur2 = preHead2;
		While [head!= null] {
			if (Head.val < x) {
				cur1.next = head;
				Cur1 = Cur1.next;
			} else {
				cur2.next = head;
				CUR2 = Cur2.next;
			}
			head = Head.next;
		}
		Cur1.next = Prehead2.next;
		Cur2.next = null;
		Return Prehead1.next

	}
}