[LintCode] 599 Insert into a Cyclic Sorted List 解题报告

Description
Given a node from a cyclic linked list which has been sorted, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be any single node in the list. Return the inserted new node.


Notice
3->5->1 is a cyclic list, so 3 is next node of 1.
3->5->1 is same with 5->1->3


Example
Given a list, and insert a value 4:
3->5->1
Return 5->1->3->4

5/18/2017

算法班，未经验证

先找到list的头，再来循环找插入点。

public class Solution {
    /**
     * @param node a list node in the list
     * @param x an integer
     * @return the inserted new list node
     */
    public ListNode insert(ListNode node, int x) {
        // Write your code here

        ListNode newNode = new ListNode(x);
        if (node == null) {
            newNode.next = newNode;
            return newNode;
        } else if (node.next == node) {
            node.next = newNode;
            newNode.next = node;
            return node;
        }
        ListNode maxNode = node;

        while (maxNode.next.val > maxNode.val) {
            maxNode = maxNode.next;
        }
        if (maxNode.val < newNode.val || maxNode.next.val > newNode.val) {
            newNode.next = maxNode.next;
            maxNode.next = newNode;
            return newNode;
        }

        ListNode node = maxNode.next;
        while (node.next.val < newNode.val) {
            node = node.next;
        }
        newNode.next = node.next;
        node.next = newNode;
        return newNode;
    }
}

但是实际上可以在单纯遍历的时候找插入点

别人的答案

public class Solution {
    /**
     * @param node a list node in the list
     * @param x an integer
     * @return the inserted new list node
     */
    public ListNode insert(ListNode node, int x) {
        // Write your code here
        
        if (node == null) {
            node = new ListNode(x);
            node.next = node;
            return node;
        }
        
        ListNode head = node;
        while (node != null && node.next != null) {
            if (node.val < node.next.val) {
                if (node.val <= x && x <= node.next.val) {
                    insertNode(node, x);
                    break;
                }
            }
            else if (node.val > node.next.val) {
                if (x > node.val || x < node.next.val) {
                    insertNode(node, x);
                    break;
                }
            }
            else { // node.val == node.next.val
                if (node.next == head) {
                    insertNode(node, x);
                    break;
                }
            }
            node = node.next;
        }
        
        return head;
    }
    
    public void insertNode(ListNode node, int x) {
        ListNode newNode = new ListNode(x);
        newNode.next = node.next;
        node.next = newNode;
    }
}