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.（Medium）
For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.
分析：
把链表归并，其思路就是先开两个链表，把小于x的值接在链表left后面，大于x的值接在链表right后面；
然后把链表left的尾部与链表right的头部接在一起。
注意：链表right的尾部next赋值为nullptr。
代码：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* partition(ListNode* head, int x) { 12 ListNode dummy1(0); 13 ListNode dummy2(0); 14 ListNode* left = &dummy1; 15 ListNode* right = &dummy2; 16 while (head != nullptr) { 17 if (head -> val < x) { 18 left -> next = head; 19 left = left -> next; 20 } 21 else { 22 right -> next = head; 23 right = right -> next; 24 } 25 head = head -> next; 26 } 27 left -> next = dummy2.next; 28 right -> next = nullptr; 29 return dummy1.next; 30 } 31 };