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.
这道题没看答案之前不明白到底如何移动，看了答案发现太简单了。new两个链表，遍历原链表，把val大于等于x的节点接到big后面，把val小于x的节点接到small后面，最后把big接到small后面，就可以了。
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
if (head == null || head.next == null){
return head;
}
ListNode big = new ListNode(-1);
ListNode small = new ListNode(-1);
ListNode bighead = big;
ListNode smallhead = small;
while (head != null){
if (head.val < x){
smallhead.next = head;
smallhead = smallhead.next;
} else {
bighead.next = head;
bighead = bighead.next;
}
head = head.next;
}
bighead.next = null;
smallhead.next = big.next;
return small.next;
}
}
