GuilinDev

Lc0203

05 August 2008

203 - Remove Linked List Elements

原题概述

Remove all elements from a linked list of integers that have value val.

Example:

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Input:  1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5

题意和分析

考察链表的基本操作，因为第一结点有可能也会被删除，所以使用dummy假头，然后返回dummy.next。

代码

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/**
 * 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);
        dummy.next = head;
        
        ListNode pre = dummy;
        
        while (pre.next != null) {
            //只要还和val相等就不要挪动指针
            if (pre.next.val == val) {
                pre.next = pre.next.next;
            } else {
                pre = pre.next; 
            }
        }
        return dummy.next;
    }
}

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class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(-1);
        ListNode pre = dummy;
        dummy.next = head;
        while (pre.next != null) {
            if (pre.next.val == val) {
                ListNode node = pre.next;
                pre.next = node.next;
                node.next = null;
            } else {
                pre = pre.next;
            }
        }
        return dummy.next;
    }
}

当然，Linked List是天然的递归结构，可以用递归来做

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if (head == null) {
            return null;
        }
        head.next = removeElements(head.next, val);
        return head.val == val ? head.next : head;//等于val就跳过现在结点（删除）
    }
}
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