题目:
题解:
/**
* 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) {
if(head == null){
return head;
}
ListNode dummy = new ListNode(-1,head);
ListNode pre = dummy;
ListNode cur = head;
while(cur != null){
if(cur.val == val){
pre.next = cur.next;
}
else {
pre = cur;
}
cur = cur.next;
}
return dummy.next;
}
}
这题的思想就是加了个虚拟头结点然后在头结点需要删除的时候更加方便
第三题
题解:
/**
* 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) {
ListNode pre = null;
ListNode cur = head;
ListNode temp = null;
while(cur != null){
temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
}
return pre;
}
}
这一题感觉也还好,这道题的思想有点像刚学C语言那会的冒泡排序
第二道题我今天事有点多就不写了,那个比较耗费精力