LeetCode 86. 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.

链表分区，把比x小的放在前面，其余的放在后面，并不改变其相对位置。

思路是建立两个链表分别存放大数和小数，再把两个链表合在一起

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head==null||head.next==null)return head;
        ListNode shead = new ListNode(0);
        ListNode lhead = new ListNode(0);
        ListNode s = shead;
        ListNode l = lhead;
        while(head!=null){
            if(head.val>=x){
                l.next = head;
                l = l.next;
            }
            else{
                s.next = head;
                s = s.next;
            }
            head = head.next;
        }
        s.next = lhead.next;
        l.next = null;
        return shead.next;
    }
}