203. 移除链表元素(简单)

1,问题描述

203. 移除链表元素

难度:简单

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:

img

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输入:head = [1,2,6,3,4,5,6], val = 6
输出:[1,2,3,4,5]

示例 2:

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输入:head = [], val = 1
输出:[]

示例 3:

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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]
  • 1 <= Node.val <= 50
  • 0 <= val <= 50

2,初步思考

​ 这个很简单,直接遍历删除、重连即可解决;然后在看到题解中有递归算法,我又立马想了递归的题解

​ 递归的解法太优雅了!!!

3,代码处理

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import support.ListNode;

public class _203移除链表元素 {

// 解法2:递归解决,先解决下一个节点,再解决当前节点
public ListNode removeElements_recursion(ListNode head, int val) {
if(head == null) return null;
head.next = removeElements_recursion(head.next, val);
return head.val == val ? head.next : head;
}

// 解法1:迭代解决,新增一个头节点,因为头节点可能也会被变动
public ListNode removeElements_iterate(ListNode head, int val) {
ListNode headNew = new ListNode(-1);
ListNode before = headNew;
before.next = head;
ListNode tail = head;
while (tail!=null){
if(tail.val == val){
before.next = tail.next;
}else {
before = before.next;
}
tail = tail.next;
}
return headNew.next;
}

public static void main(String[] args) {
_203移除链表元素 removeElements = new _203移除链表元素();
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(6);
head.next.next.next = new ListNode(3);
head.next.next.next.next = new ListNode(4);
head.next.next.next.next.next = new ListNode(5);
head.next.next.next.next.next.next = new ListNode(6);
// ListNode listNode = removeElements.removeElements(head, 6);
ListNode listNode = removeElements.removeElements_recursion(head, 6);
while (listNode!=null){
System.out.println(listNode.val);
listNode = listNode.next;
}
}
}

下面是官方的题解

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class Solution {
public ListNode removeElements_gov_recursion(ListNode head, int val) {
if (head == null) {
return head;
}
head.next = removeElements_gov_recursion(head.next, val);
return head.val == val ? head.next : head;
}

public ListNode removeElements_gov_iterate(ListNode head, int val) {
ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode temp = dummyHead;
while (temp.next != null) {
if (temp.next.val == val) {
temp.next = temp.next.next;
} else {
temp = temp.next;
}
}
return dummyHead.next;
}

}