LeetCode86 Partition List

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：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* partition(ListNode* head, int x) {
         ListNode dummy1();
         ListNode dummy2();
         ListNode* left = &dummy1;
         ListNode* right = &dummy2;
         while (head != nullptr) {
             if (head -> val < x) {
                 left -> next = head;
                 left = left -> next;
             }
             else {
                 right -> next = head;
                 right = right -> next;
             }
             head = head -> next;
         }
         left -> next = dummy2.next;
         right -> next = nullptr;
         return dummy1.next;
     }
 };