Leetcode-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.

Analysis:

The first element that is not less than x is the point of partition. We record this node and its predecessor. Once we encount a node that is less than x later, we insert it between the point and its predecessor, then update the predecssor to this node.

Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head==null || head.next==null) return head;
        
        ListNode preHead = new ListNode(0);
        preHead.next = head;
        ListNode end = preHead;
        ListNode cur = head;
        
        while (cur!=null && cur.val<x){
            end = cur;
            cur = cur.next;
        }
        
        while (cur!=null && cur.next!=null){
            if (cur.next.val>=x){
                cur = cur.next;
            } else {
                ListNode temp = cur.next;
                cur.next = temp.next;
                temp.next = end.next;
                end.next = temp;
                end = temp;
            }
        }
        
        return preHead.next;
    }
}