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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

题目大意：

将LinkedList利用partition划分为两个部分，LinkedList的前半部分都是比给定的x要小的值，而后半部分是比给定值要大的值。

解法：

将该链表一分为二，分成两个链表，最后再将两个链表串起来。链表一存放的都是比value值要小的节点，而链表二存放的都是比value值要大的节点。

java:

class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode h1=new ListNode(0);
        ListNode h2=new ListNode(0);
        ListNode head1=h1,head2=h2;
        while(head!=null){
            if(head.val<x){
                head1.next=head;
                head1=head1.next;
            }else{
                head2.next=head;
                head2=head2.next;
            }
            head=head.next;
        }
        head2.next=null;
        head1.next=h2.next;

        return h1.next;
    }
}