【Leetcode】【Medium】Partition List

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.

解题思路：

1、先条件：输入链表不为空；

2、表头可能改变，因此需要新建一个结点指向表头，或使用二维指针；

3、不变参数：

　　curNode永远指向待操作结点的前驱（方便删减）

　　small_end永远指向已经分割的所有小结点中，最后一个结点

　　big_begin永远指向已经分割的所有大结点中，第一个结点

　　small_end->next = big_begin;

4、curNode->next为NULL时，循环结束。当发现小结点时，插入small_end和big_begin中间；

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) {
         if (head == NULL)
             return head;
         ListNode* prehead = new ListNode();
         prehead->next = head;
         ListNode* curNode = prehead;
         ListNode* small_end = NULL;
         ListNode* big_begin = NULL;

         while (curNode->next && curNode->next->val < x)
             curNode = curNode->next;

         small_end = curNode;
         big_begin = small_end->next;

         while (curNode->next) {
             if (curNode->next->val < x) {
                 small_end->next = curNode->next;
                 small_end = small_end->next;
                 curNode->next = curNode->next->next;
                 small_end->next = big_begin;
             } else {
                 curNode = curNode->next;
             }
         }

         head = prehead->next;
         delete prehead;
         return head;
     }
 };