本文已收录到:LeetCode刷题 专题
- 42. 接雨水 (Trapping Rain Water)
- [LeetCode] 2. 两数相加 2. Add Two Numbers
- [LeetCode] 19. 删除链表的倒数第N个节点 19. Remove Nth Node From End of List
- [LeetCode] 21. 合并两个有序链表(Merge Two Sorted Lists)
- [LeetCode] 24. 两两交换链表中的节点(Swap Nodes in Pairs)
- [LeetCode] 61. 旋转链表(Rotate List)
- [LeetCode] 82. 删除排序链表中的重复元素 II(Remove Duplicates from Sorted List II)
- [LeetCode] 83. 删除排序链表中的重复元素(Remove Duplicates from Sorted List)
- [LeetCode] 86. 分隔链表(Partition List)
- [LeetCode] 206. 反转链表(Reverse Linked List)
- [LeetCode] 92. 反转链表 II(Reverse Linked List II)
- [LeetCode] 147. 对链表进行插入排序(Insertion Sort List)
- [LeetCode] 203. 移除链表元素(Remove Linked List Elements)
- [LeetCode] 160. 相交链表(Intersection of Two Linked Lists)
- [LeetCode] 237. 删除链表中的节点 (Delete Node in a Linked List)
- [LeetCode] 328. 奇偶链表 (Odd Even Linked List)
- [LeetCode] 876. 链表的中间结点 (Middle of the Linked List)
- [LeetCode] 1290. 二进制链表转整数 (Convert Binary Number in a Linked List to Integer)
- [LeetCode] 141. 环形链表 (Linked List Cycle)
- [LeetCode] 142. 环形链表 II (Linked List Cycle II)
- [LeetCode] 283. 移动零 (Move Zeroes)
- [LeetCode] 496. 下一个更大元素 I (Next Greater Element I)
- [LeetCode] 503. 下一个更大元素 II (Next Greater Element II)
- [LeetCode] 456. 132模式 (132 Pattern)
- [LeetCode] 20. 有效的括号 (Valid Parentheses)
- [LeetCode] 94. 二叉树的中序遍历 (Binary Tree Inorder Traversal)
- [LeetCode] 144. 二叉树的前序遍历 (Binary Tree Preorder Traversal)
- [LeetCode] 145. 二叉树的后序遍历 (Binary Tree Postorder Traversal)
[title]题目[/title]
给定一个链表和一个特定值 x,对链表进行分隔,使得所有小于 x 的节点都在大于或等于 x 的节点之前。
你应当保留两个分区中每个节点的初始相对位置。
示例:
输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/partition-list
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
[title]视频讲解[/title]
[bilibili cid=”” page=”1″]582631392[/bilibili]
[title]代码[/title]
- /**
- * Definition for singly-linked list.
- * struct ListNode {
- * int val;
- * ListNode *next;
- * ListNode(int x) : val(x), next(NULL) {}
- * };
- */
- class Solution {
- public:
- ListNode* partition(ListNode* head, int x) {
- ListNode dummy1(0);
- ListNode* cur1 = &dummy1;
- ListNode dummy2(0);
- ListNode* cur2 = &dummy2;
- while (head != nullptr)
- {
- if (head->val < x)
- {
- cur1->next = head;
- cur1 = head;
- }
- else
- {
- cur2->next = head;
- cur2 = head;
- }
- head = head->next;
- }
- cur1->next = dummy2.next;
- cur2->next = nullptr;
- return dummy1.next;
- }
- };