Delete Node in a Linked List

最新推荐文章于 2019-07-28 09:29:27 发布

原创最新推荐文章于 2019-07-28 09:29:27 发布 · 296 阅读

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本文介绍了一种在O(1)时间复杂度内删除单链表中指定节点的方法，避免了传统遍历查找的低效，直接通过修改节点值和指针实现快速删除，适用于非表头或表尾的任意节点。

372. Delete Node in a Linked List

给定一个单链表中的一个等待被删除的节点(非表头或表尾)。请在在O(1)时间复杂度删除该链表节点。

样例

Linked list is 1->2->3->4, and given node 3, delete the node in place 1->2->4

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */


class Solution {
public:
    /*
     * @param node: the node in the list should be deletedt
     * @return: nothing
     */
    void deleteNode(ListNode * node) {
        // write your code here
        if(node==NULL) return;
        if(node->next==NULL)
        {
            node=NULL;
            return;
        }
        node->val=node->next->val;
        node->next=node->next->next;
    }
};


###################################################

"""
Definition of ListNode
class ListNode(object):

    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""


class Solution:
    """
    @param: node: the node in the list should be deletedt
    @return: nothing
    """
    def deleteNode(self, node):
        # write your code here
        if node==None:
            return None
        if node.next==None:
            return None 
        node.val=node.next.val
        node.next=node.next.next
        return node