LeetCode#237. Delete Node in a Linked List
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- 题目:删除链表中的某个节点
- 难度:Easy
- 思路:分三种情况
- 待删除节点不是尾节点:将待删除节点的下一个节点的值赋给待删除节点,然后待删除节点的指向下一个节点的指针调整指向下下个节点
- 待删除节点也就是头结点(链表只有一个节点):将头结点和待删除节点都赋值为null
- 待删除节点是尾节点:找到待删除节点的前一个节点,将待删除节点赋值为null,将前一个节点的next指针指向null
- 代码:
private void deleteOneNode(ListNode head, ListNode toBeDelete){ if(head == null || toBeDelete == null){ return ; } //不是尾节点:将待删除节点的下一个节点的值赋给待删除节点,然后删除下一个节点 if(toBeDelete.next != null){ toBeDelete.val = toBeDelete.next.val; toBeDelete.next = toBeDelete.next.next; }else if(head == toBeDelete){ //链表只有一个节点 head = null; toBeDelete = null; }else{ //是尾节点:找到倒数第二个节点 while(head.next != toBeDelete){ head = head.next; } toBeDelete = null; head.next = toBeDelete; } }
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