【问题描述】[中等]

给定单向链表的头指针和一个要删除的节点的值，定义一个函数删除该节点。返回删除后的链表的头节点。注意：此题对比原题有改动示例 1:输入: head = [4,5,1,9], val = 5
输出: [4,1,9]
解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.

【解答思路】

1. 双指针

时间复杂度：O(N^2) 空间复杂度：O(1)

class Solution {public ListNode deleteNode(ListNode head, int val) {if(head.val == val) return head.next;ListNode pre = head, cur = head.next;while(cur != null && cur.val != val) {pre = cur;cur = cur.next;}if(cur != null) pre.next = cur.next;return head;}
}

2.单指针

时间复杂度：O(N) 空间复杂度：O(1)

 public ListNode deleteNode(ListNode head, int val) {if(head.val == val) return head.next;ListNode cur = head;while(cur != null){if(cur.next.val == val ){cur.next = cur.next.next;break;}cur=cur.next;}return head;}

【总结】

1.画图理解指针问题

2.遍历指针

 ListNode cur = head;while(cur != null)cur=cur.next;}

3.链表为空作判断 链表头尾特殊判断

参考链接：https://leetcode-cn.com/problems/shan-chu-lian-biao-de-jie-dian-lcof/solution/mian-shi-ti-18-shan-chu-lian-biao-de-jie-dian-sh-2/