86.Partition List (M)

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

1.Description(Easy)

Given the head of 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.

Example 1:

Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]

Example 2:

Input: head = [2,1], x = 2
Output: [1,2]

Constraints:

• The number of nodes in the list is in the range [0, 200].

• -100 <= Node.val <= 100

• -200 <= x <= 200

2.Code

创建两个左右链表分别连接比X小和大的节点，最后把这两个连接起来。

 public ListNode partition(ListNode head, int x) {
        ListNode leftDummy=new ListNode(0);
        ListNode rightDummy=new ListNode(0);
        ListNode left=leftDummy;
        ListNode right=rightDummy;

        while(head!=null){
            if(head.val<x){
                left.next=head;
                left=head;//be equal to left=left.next;
            }
            else{
                right.next=head;
                right=head;
            }
            head=head.next;
        }

        right.next=null;
        left.next=rightDummy.next;
       return leftDummy.next;       
    }