题目：
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
算法解析：
申请新表头 NewHead1,NewHead2, 把链表中小于x节点连接在p1后面，其余节点连接在p2后面。链表遍历完成后，将链表1的尾部与链表2的头部连接
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode() : val(0), next(nullptr) {}* ListNode(int x) : val(x), next(nullptr) {}* ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* partition(ListNode* head, int x) {ListNode* NewHead1 = new ListNode(0); ListNode* NewHead2 = new ListNode(0);ListNode* p1 = NewHead1, *p2 = NewHead2,*phead = head;while(phead){if(phead->val < x){p1->next = phead;p1 = p1->next;}else{p2->next = phead;p2 = p2->next;}phead = phead->next;}p2->next = NULL;p1->next = NewHead2->next;return NewHead1->next;}
};