LeetCode之“链表”：Partition List

　　题目链接

　　题目要求：

　　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.

　　该题利用dummy节点能极大方便编程，具体程序如下：

 /**
  * 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) {
         if(!head || !head->next)
             return head;

         ListNode *dummy = new ListNode(INT_MIN);
         dummy->next = head;
         ListNode *parNode = dummy;
         ListNode *preNode = nullptr, *curNode = nullptr;
         while(parNode && parNode->val < x)
         {
             preNode = parNode;
             parNode = parNode->next;
         }

         parNode = preNode;
         if(!parNode || !parNode->next)
         {
             head = dummy->next;
             delete dummy;
             dummy = nullptr;
             return head;
         }

         preNode = parNode->next;
         curNode = preNode->next;
         while(curNode)
         {
             if(curNode->val < x)
             {
                 ListNode *nextPar = parNode->next, *nextCur = curNode->next;
                 parNode->next = curNode;
                 curNode->next = nextPar;
                 parNode = parNode->next;

                 curNode = nextCur;
                 preNode->next = curNode;
             }
             else
             {
                 preNode = preNode->next;
                 curNode = curNode->next;
             }
         }

         head = dummy->next;
         delete dummy;
         dummy = nullptr;

         return head;
     }
 };