86. Partition List

Posted 2020-08-09 三颗心

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.

 

使用两个链表newHead1, newHead2，遍历原链表，如果结点值小于x，则挂在newHead1上，如果大于等于x，则挂在newHead2上，最后把newHead2挂在newHead1上。

 

 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         ListNode* newHead1 = new ListNode(0);
13         ListNode* newHead2 = new ListNode(0);
14         
15         ListNode* pNode1 = newHead1;
16         ListNode* pNode2 = newHead2;
17         
18         ListNode* pNode = head;
19         
20         while(pNode){
21             if(pNode->val < x){
22                 pNode1->next = pNode;
23                 pNode1 = pNode1->next;
24             }else{
25                 pNode2->next = pNode;
26                 pNode2 = pNode2->next;
27             }
28             pNode = pNode->next;
29         }
30         
31         pNode2->next = NULL;
32         pNode1->next = newHead2->next;
33         
34         return newHead1->next;
35     }
36 };