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Leetcode | Partition List

2024-07-04 20:06:32   227人阅读

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.

创建两个链表，一个比x小，一个大于等于x。最后把它们连起来。注意把第二个链表最后一个元素置为NULL。不然可能产生死循环。

 1 class Solution {
 2 public:
 3     ListNode *partition(ListNode *head, int x) {
 4         ListNode n1(0), n2(0);
 5         ListNode *p1 = &n1, *p2 = &n2;
 6         
 7         while (head != NULL) {
 8             if (head->val < x) {
 9                 p1->next = head;
10                 p1 = p1->next;
11             } else {
12                 p2->next = head;
13                 p2 = p2->next;
14             }
15             head = head->next;
16         }
17         p2->next = NULL;
18         p1->next = n2.next;
19         return n1.next;
20     }
21 };

这里n1和n2不用new，省得最后还要delete.


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