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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
难度：medium
题目：给定链表和一指定值Ｘ，以Ｘ为参照将链表划分为小于Ｘ和大于等于Ｘ两部分。保持原始链表中结点的顺序不变。
思路：将原始链表先拆分为两链表，一个收集所以小于Ｘ的，一个收集其它的。然后合并两链表。
Runtime: 0 ms, faster than 100.00% of Java online submissions for Partition List.
Memory Usage: 36.7 MB, less than 0.90% of Java online submissions for Partition List.
/**
* 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) {
ListNode lHead = new ListNode(0), lTail = lHead;
ListNode geHead = new ListNode(0), geTail = geHead;
ListNode ptr = head, node = null;
while (ptr != null) {
node = ptr;
ptr = ptr.next;
if (node.val < x) {
lTail.next = node;
lTail = node;
} else {
geTail.next = node;
geTail = node;
}
}
lTail.next = geHead.next;
geTail.next = null;
return lHead.next;
}
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