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Remove all elements from a linked list of integers that have value val.
Example Given: 1 -- > 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6 Return: 1 -- > 2 --> 3 --> 4 --> 5
Credits: Special thanks to @mithmatt for adding this problem and creating all test cases.
这道移除链表元素是链表的基本操作之一,没有太大的难度,就是考察了基本的链表遍历和设置指针的知识点,我们只需定义几个辅助指针,然后遍历原链表,遇到与给定值相同的元素,将该元素的前后连个节点连接起来,然后删除该元素即可,要注意的是还是需要在链表开头加上一个dummy node,具体实现参见代码如下:
解法一:
class Solution { public: ListNode* removeElements(ListNode* head, int val) { ListNode *dummy = new ListNode(-1), *pre = dummy; dummy->next = head; while (pre->next) { if (pre->next->val == val) { ListNode *t = pre->next; pre->next = t->next; t->next = NULL; delete t; } else { pre = pre->next; } } return dummy->next; } };
如果只是为了通过OJ,不用写的那么严格的话,下面这种方法更加简洁,当判断下一个结点的值跟给定值相同的话,直接跳过下一个结点,将next指向下下一个结点,而根本不断开下一个结点的next,更不用删除下一个结点了。最后还要验证头结点是否需要删除,要的话直接返回下一个结点,参见代码如下:
解法二:
class Solution { public: ListNode* removeElements(ListNode* head, int val) { if (!head) return NULL; ListNode *cur = head; while (cur->next) { if (cur->next->val == val) cur->next = cur->next->next; else cur = cur->next; } return head->val == val ? head->next : head; } };
我们也可以用递归来解,写法很简洁,通过递归调用到链表末尾,然后回来,需要要删的元素,将链表next指针指向下一个元素即可:
解法三:
class Solution { public: ListNode* removeElements(ListNode* head, int val) { if (!head) return NULL; head->next = removeElements(head->next, val); return head->val == val ? head->next : head; } };
类似题目:
Remove Element
Delete Node in a Linked List
参考资料:
https://leetcode.com/problems/remove-linked-list-elements/
https://leetcode.com/problems/remove-linked-list-elements/discuss/57324/AC-Java-solution
https://leetcode.com/problems/remove-linked-list-elements/discuss/57306/3-line-recursive-solution
https://leetcode.com/problems/remove-linked-list-elements/discuss/57331/Accepted-7-line-clean-java-solution
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered:
解法三的递归代码没有delete
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Remove all elements from a linked list of integers that have value val.
Example
Given: 1 -- > 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 -- > 2 --> 3 --> 4 --> 5
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
这道移除链表元素是链表的基本操作之一,没有太大的难度,就是考察了基本的链表遍历和设置指针的知识点,我们只需定义几个辅助指针,然后遍历原链表,遇到与给定值相同的元素,将该元素的前后连个节点连接起来,然后删除该元素即可,要注意的是还是需要在链表开头加上一个dummy node,具体实现参见代码如下:
解法一:
如果只是为了通过OJ,不用写的那么严格的话,下面这种方法更加简洁,当判断下一个结点的值跟给定值相同的话,直接跳过下一个结点,将next指向下下一个结点,而根本不断开下一个结点的next,更不用删除下一个结点了。最后还要验证头结点是否需要删除,要的话直接返回下一个结点,参见代码如下:
解法二:
我们也可以用递归来解,写法很简洁,通过递归调用到链表末尾,然后回来,需要要删的元素,将链表next指针指向下一个元素即可:
解法三:
类似题目:
Remove Element
Delete Node in a Linked List
参考资料:
https://leetcode.com/problems/remove-linked-list-elements/
https://leetcode.com/problems/remove-linked-list-elements/discuss/57324/AC-Java-solution
https://leetcode.com/problems/remove-linked-list-elements/discuss/57306/3-line-recursive-solution
https://leetcode.com/problems/remove-linked-list-elements/discuss/57331/Accepted-7-line-clean-java-solution
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered: