【Lintcode】096.Partition List
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; } };
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步