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;

     }

 };

巴特西

LeetCode86 Partition List

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