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.
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} x
* @return {ListNode}
*/
var partition = function(head, x) {
if (head === null) {
return null;
}
var cur = head;
var big = new ListNode();
var big_cur = big;
while (cur !== null && cur.val >= x) {
big_cur.next = cur;
big_cur = big_cur.next;
cur = cur.next;
}
if (cur === null) {
return head;
}
head = cur;
var pre = head;
cur = head.next;
while (cur !== null) {
if (cur.val < x) {
pre = cur;
cur = cur.next;
} else {
big_cur.next = cur;
big_cur = big_cur.next;
cur = cur.next;
pre.next = cur;
}
}
big_cur.next = null;
pre.next = big.next;
return head;
};