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237. Delete Node in a Linked List
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
Given linked list — head = [4,5,1,9], which looks like following:
Example 1:
Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.Example 2:
Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.Note:
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The linked list will have at least two elements.
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All of the nodes' values will be unique.
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The given node will not be the tail and it will always be a valid node of the linked list.
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Do not return anything from your function.
思路分析
这个题其实很简单！把节点的值覆盖当前节点的值即可。
没想到打脸如此之快！还有更简单的办法，两行代码搞定：①把下一个节点的值拷贝到当前节点；②把当前节点的下一节点指向下下一个节点即可。
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一刷
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二刷
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/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Delete Node in a Linked List.
*
* Memory Usage: 36.2 MB, less than 100.00% of Java online submissions for Delete Node in a Linked List.
*
* Copy from: https://leetcode.com/problems/delete-node-in-a-linked-list/solution/[Delete Node in a Linked List solution - LeetCode]
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-13 20:28
*/
public void deleteNode(ListNode node) {
node.val = node.next.val;
node.next = node.next.next;
}
/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Delete Node in a Linked List.
*
* Memory Usage: 36.5 MB, less than 100.00% of Java online submissions for Delete Node in a Linked List.
*/
public void deleteNodeCopyList(ListNode node) {
ListNode current = node;
while (Objects.nonNull(current) && Objects.nonNull(current.next)) {
ListNode next = current.next;
current.val = next.val;
if (Objects.nonNull(next.next)) {
current = next;
} else {
current.next = null;
}
}
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-16 20:54:33
*/
public void deleteNode(ListNode node) {
// 把后面一个元素拷贝过来
node.val = node.next.val;
// 把当前元素的下节点指针指向下下节点
node.next = node.next.next;
}