代码随想录day04

最新推荐文章于 2024-12-27 16:17:53 发布

原创最新推荐文章于 2024-12-27 16:17:53 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#链表 #数据结构 #java

代码随想录刷题记录专栏收录该内容

36 篇文章

订阅专栏

本文详细解析了四个经典的链表算法题目：两两交换链表中的节点、删除链表的倒数第N个节点、链表相交及环形链表II。提供了包括设置虚拟头结点、双指针等技巧在内的多种解决方案。

一、两两交换链表中的节点（LeetCode24）

设置虚拟头结点

/**
 * 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) {
        ListNode dummyHead = new ListNode(-1,head); //设置虚拟头结点
        dummyHead.next = head;
        ListNode currentHead = dummyHead; //设置临时指针节点
        ListNode tem = null,tem1 = null; //设置临时保存节点
        while(currentHead.next!=null&&currentHead.next.next!=null){
            tem = currentHead.next;
            tem1 = currentHead.next.next.next;
            currentHead.next = currentHead.next.next;
            currentHead.next.next = tem;
            tem.next = tem1;
            currentHead = currentHead.next.next; //相邻两个节点交换位置后，指针后移两位
        }
        return dummyHead.next;
    }
}

递归

/**
 * 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 next = head.next; //获取当前节点的下一个节点
        ListNode newNode = swapPairs(next.next); //进行递归
        //进行交换
        next.next = head;
        head.next = newNode;
        return next;
    }
}

二、删除链表的倒数第N个节点（LeetCode19）

双指针

/**
 * 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 dummyHead = new ListNode(-1,head);
        ListNode fast = dummyHead;
        ListNode slow = dummyHead;
        //快慢指针相差n+1个节点
        n++;
        while(n-- > 0){
            fast = fast.next;    
        }
        while(fast != null){
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return dummyHead.next;
    }
}

三、链表相交（面试题 02.07）

设置双指针，链表尾端对齐后判断值相等

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode currentA = headA;
        ListNode currentB = headB;
        int lengthA = 0, lengthB = 0;
        while (currentA != null) { // 求链表A的长度
            lengthA++;
            currentA = currentA.next;
        }
        while (currentB != null) { // 求链表B的长度
            lengthB++;
            currentB = currentB.next;
        }
        //让当前指针重新回到头结点
        currentA = headA;
        currentB = headB;
        //使其A为长链表
        if(lengthA < lengthB){
            //长度交换
            int temLength = lengthA;
            lengthA = lengthB;
            lengthB = temLength;
            //链表交换
            ListNode newlist = currentA;
            currentA = currentB;
            currentB = newlist;
        }
        //末尾对齐，长链表当前指针移动与短链表指针对齐
        for(int i = 0;i < lengthA - lengthB;i++){
            currentA = currentA.next;
        }
        //循环找出公共点
        while(currentA != null){
            if(currentA == currentB){
                return currentA;
            }
            currentA = currentA.next;
            currentB = currentB.next;
        }
        return null;
    }
}

四、环形链表II（LeetCode142）

双指针，判断环后再次设置双指针相遇即为入口

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
            if(fast == slow){ //有环
                ListNode index1 = fast;
                ListNode index2 = head;
                // 两个指针，从头结点和相遇结点，各走一步，直到相遇，相遇点即为环入口
                while(index1 != index2){
                    index1 = index1.next;
                    index2 = index2.next;
                }
                return index1;
            }
        }
        return null;
    }
}