LCR.P29[循环有序列表的插入]

方法一：模拟

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/*
// Definition for a Node.
class Node {
    public int val;
    public Node next;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _next) {
        val = _val;
        next = _next;
    }
};
*/

class Solution {
    public Node insert(Node head, int insertVal) {
        Node node = new Node(insertVal);
        if (head == null) {
            node.next = node;
            return node;
        }
        if (head.next == head) {
            head.next = node;
            node.next = head;
            return head;
        }
        Node cur = head, next = head.next;
        while (next != head) {
            if (insertVal >= cur.val && insertVal <= next.val) break;
            if (cur.val > next.val) {
                // 此时 cur 为循环链表的最大值， next 为循环列表的最小值
                if (insertVal > cur.val || insertVal < next.val) break;
            }
            cur = cur.next;
            next = next.next;
        }
        cur.next = node;
        node.next = next;
        return head;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(1)$