0237. Delete Node in a Linked List

237. Delete Node in a Linked List #

题目 #

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:

  • The linked list will have at least two elements.
  • All of the nodes’ values will be unique.
  • The given node will not be the tail and it will always be a valid node of the linked list.
  • Do not return anything from your function.

题目大意 #

删除给点结点。没有给链表的头结点。

解题思路 #

其实就是把后面的结点都覆盖上来即可。或者直接当前结点的值等于下一个结点,Next 指针指向下下个结点,这样做也可以,只不过中间有一个结点不被释放,内存消耗多一些。

代码 #


package leetcode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteNode(node *ListNode) {
	if node == nil {
		return
	}
	cur := node
	for cur.Next.Next != nil {
		cur.Val = cur.Next.Val
		cur = cur.Next
	}
	cur.Val = cur.Next.Val
	cur.Next = nil
}


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