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.
根据给出的链表和基准值，把链表分为两部分，比基准值小的在前，大的在后，两部分中的元素分别保持原来在链表中的顺序。
很直接的实现就好，构造出左右两个链表，然后拼起来：
var partition = function(head, x) {
var leftHead = null;
var left = null;
var rightHead = null;
var right = null;
var next = null;
while (head) {
next = head.next;
if (head.val<x) {
if (leftHead) left.next = head;
else leftHead = head;
left = head;
head.next = null;
} else {
if (rightHead) right.next = head;
else rightHead = head;
right = head;
head.next = null;
}
head = next;
}
if (left) {
left.next = rightHead;
return leftHead;
} else {
return rightHead;
}
};
