描述：
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.
代码：
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
ListNode left_dummy(-1); //头结点
ListNode right_dummy(-1); //头结点
ListNode *left_cur = &left_dummy;
ListNode *right_cur = &right_dummy;
for (ListNode *cur = head; cur; cur = cur->next) {
if (cur->val < x) {
left_cur->next = cur;
left_cur = cur;
}
else {
right_cur->next = cur;
right_cur = cur;
}
}
left_cur->next = right_dummy.next;
right_cur->next = NULL;
return left_dummy.next;
}
};