【LeetCode 86】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.

思路：

　　设置两个链表头，遍历原链表，一个追加小数链表，一个追加大数链表，最后将小数链表粘到大数链表前边即为结果。

C++：

 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         if(head == 0)
13             return 0;
14
15         ListNode* pless = 0;
16         ListNode* plesshead = 0;
17         ListNode* pgreat = 0;
18         ListNode* pgreathead = 0;
19         ListNode* phead = head;
20
21         while(phead != 0)
22         {
23             if(phead->val < x)
24             {
25                 if(pless == 0)
26                 {
27                     pless = phead;
28                     plesshead = pless;
29                 }
30                 else
31                 {
32                     pless->next = phead;
33                     pless = pless->next;
34                 }
35             }
36             else
37             {
38                 if(pgreat == 0)
39                 {
40                     pgreat = phead;
41                     pgreathead = pgreat;
42                 }
43                 else
44                 {
45                     pgreat->next = phead;
46                     pgreat = pgreat->next;
47                 }
48             }
49
50             phead = phead->next;
51         }
52
53         if(plesshead == 0)
54             return pgreathead;
55
56         if(pgreathead == 0)
57             return plesshead;
58
59         pless->next = pgreat->next = 0;
60         pless->next = pgreathead;
61
62         return plesshead;
63     }
64 };