【随想】每日两题Day.6

简介: 【随想】每日两题

题目:LeetCode 24.两两交换链表中的结点

给你一个链表,两两交换其中相邻的节点,并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题(即,只能进行节点交换)。

示例 1:

输入:head = [1,2,3,4]

输出:[2,1,4,3]

示例 2:


输入:head = []

输出:[]

示例 3:


输入:head = [1]

输出:[1]

提示:


链表中节点的数目在范围 [0, 100] 内

0 <= Node.val <= 100

代码:

/**
 * 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 swapPairs(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode myhead = new ListNode(-1);
        ListNode tmp = myhead;
        ListNode prev = head;
        ListNode cur = prev.next;
        ListNode last = cur.next;
        while(last != null) {
            tmp.next = cur;
 
            cur.next = prev;
            tmp = prev;
            prev = last;
            cur = prev.next;
            if (cur == null) break;
            last = cur.next;
        }
        if (cur != null) {
            tmp.next = cur;
            cur.next = prev;
            prev.next = null;
        } else {
            tmp.next = prev;
        }
        return myhead.next;
    }
}

思考:

这个题的主要思路就是,用一个虚拟头结点把所有结点重新连接起来。用prev、cur、last 这三个结点,分别存要交换的两个结点和交换后的next。这样当last为空时,或者cur为空时(结点数为单数)做单独处理。此题不难理解。

题目:LeetCode 19.删除链表的导数第N个结点

给你一个链表,删除链表的倒数第 n 个结点,并且返回链表的头结点。

示例 1:

输入:head = [1,2,3,4,5], n = 2

输出:[1,2,3,5]

示例 2:


输入:head = [1], n = 1

输出:[]

示例 3:


输入:head = [1,2], n = 1

输出:[1]

提示:


链表中结点的数目为 sz

1 <= sz <= 30

0 <= Node.val <= 100

1 <= n <= sz

代码:

/**
 * 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 removeNthFromEnd(ListNode head, int n) {
        ListNode dummyNode = new ListNode(-1);
        dummyNode.next = head;
        int count = 0;
        ListNode prev = dummyNode;
        ListNode cur = dummyNode;
        while(n > 0) {
            cur = cur.next;
            n--;
        }
        while(cur.next != null) {
            cur = cur.next;
            prev = prev.next;
        }
        prev.next = prev.next.next;
        return dummyNode.next;
    }
}

思考:

对于链表操作的题,还是用虚拟头结点更简单,因为方便操作实际头结点,不需要单独处理实际头结点了。这道题用到的是快慢指针法,先让 cur 走 n 步,然后prev 和 cur 一起走,考虑好边界就找到了倒数第 n 个结点的前一个结点,最后删除。

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