题目

给定单向链表的头指针以及待删除的指针,定义一个函数在 O(1) 的时间复杂度下删除

解题思路

  1. 待删除节点非尾节点,将后一个节点的值复制到当前节点,然后删除后一个节点
  2. 待删除节点为尾节点,从头节点开始,找到待删除节点的前一个节点进行删除
public void O1DeleteNode(ListNode head, ListNode needDelete) {

    if (needDelete.next != null) {
        ListNode next = needDelete.next.next;
        needDelete.val = needDelete.next.val;
        needDelete.next = next;

    } else {
        ListNode cursor = head;

        while (cursor != null) {
            if (cursor.next == needDelete) break;

            cursor = cursor.next;
        }

        if (cursor == null) return;

        cursor.next = needDelete.next;
    }

}

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