【Leetcode】203. Remove Linked List Elements
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思路:
(1)若链表为空,则返回null。
(2)遍历整个链表,cur指向当前判断的节点,pre指向其前驱结点。
(3)若cur的值等于指定值,若cur指向的是头结点,则head指向cur的下一节点,pre指向cur;否则,pre的next指向cur的next。
(4)cur指向下一节点。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */public class Solution { public ListNode removeElements(ListNode head, int val) { if (head == null) return null; ListNode pre = null, cur = head; while (cur != null) { if (cur.val == val) { if (cur == head) { head = cur.next; pre = cur; } else pre.next = cur.next; } else pre = cur; cur = cur.next; } return head; }}Runtime:2ms
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