题目:LeetCode 203.移除列表元素
给你一个链表的头节点 head
和一个整数 val
,请你删除链表中所有满足 Node.val == val
的节点,并返回 新的头节点 。
示例 1:
输入:head = [1,2,6,3,4,5,6], val = 6 输出:[1,2,3,4,5]
示例 2:
输入:head = [], val = 1 输出:[]
示例 3:
输入:head = [7,7,7,7], val = 7 输出:[]
提示:
- 列表中的节点数目在范围
[0, 104]
内 1 <= Node.val <= 50
0 <= val <= 50
代码:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
while(head!=null && head.val==val) {
head = head.next;
}
if(head == null) return null;
ListNode cur = head.next;
ListNode prev = head;
while(cur != null) {
if(cur.val == val) {
prev.next = cur.next;
} else {
prev = cur;
}
cur = cur.next;
}
return head;
}
}
思考:
这道题很简单,注意边界和判空吧。
题目:206.反转链表
示例 1:
输入:head = [1,2,3,4,5] 输出:[5,4,3,2,1]
示例 2:
输入:head = [1,2] 输出:[2,1]
示例 3:
输入:head = [] 输出:[]
提示:
- 链表中节点的数目范围是
[0, 5000]
-5000 <= Node.val <= 5000
代码:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if(head == null || head.next == null) return head;
ListNode newhead = null;
ListNode cur = head;
//头插新节点
while(cur != null) {
ListNode tmp = cur;
cur = cur.next;
tmp.next = newhead;
newhead = tmp;
}
return newhead;
}
}
思考:
这个也很简单,头插法既简单又好理解。根据尾插理解头插,尾插就是找个新的尾,头插就是找个新的头。等二刷的时候再看什么双指针,递归法吧。