Lintcode096.Partition List
Posted Vincent丶
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Lintcode096.Partition List相关的知识,希望对你有一定的参考价值。
题目:
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.
Example
Given 1->4->3->2->5->2->null
and x = 3
,
return 1->2->2->4->3->5->null
.
题解:
Solution 1 ()
class Solution { public: ListNode *partition(ListNode *head, int x) { ListNode* left = new ListNode(-1); ListNode* right = new ListNode(-1); ListNode* l = left; ListNode* r = right; while (head) { if (head->val < x) { l->next = head; l = l->next; } else { r->next = head; r = r->next; } head = head->next; } l->next = right->next; r->next = nullptr; return left->next; } };
以上是关于Lintcode096.Partition List的主要内容,如果未能解决你的问题,请参考以下文章
lintcode378- Convert Binary Search Tree to Doubly Linked Lis- medium
[LintCode] Longest Increasing Subsequence
Lintcode077.Longest Common Subsequence
LintCode刷题笔记--Longest Increasing Subsequence