摘要：新建两个链表，分别存和的结点。令头结点分别叫作和，对应的指针分别叫作和。然后遍历，当小于的时候放入，否则放入。最后，让较小值链表尾结点指向较大值链表头结点，再让较大值链表尾结点指向。
Problem
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.
ExampleGiven 1->4->3->2->5->2->null and x = 3,
return 1->2->2->4->3->5->null.
新建两个链表，分别存
public class Solution {
public ListNode partition(ListNode head, int x) {
ListNode dummyleft = new ListNode(0);
ListNode dummyright = new ListNode(0);
ListNode left = dummyleft, right = dummyright;
while (head != null) {
if (head.val < x) {
left.next = head;
left = left.next;
}
else {
right.next = head;
right = right.next;
}
head = head.next;
}
left.next = dummyright.next;
right.next = null;
return dummyleft.next;
}
}
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