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]
给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。
示例 1:
输入: 1->1->2
输出: 1->2
示例 2:
输入: 1->1->2->3->3
输出: 1->2->3
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/remove-duplicates-from-sorted-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″]497649929[/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* deleteDuplicates(ListNode* head) {
ListNode* cur = head;
while (cur->next != nullptr && cur != nullptr)
{
if (cur->val == cur->next->val)
{
cur->next = cur->next->next;
}
else
{
cur = cur->next;
}
}
return head;
}
};
