链表题 剑指21.删除链表的倒数第n个节点剑指22.链表的倒数第k个节点234.回文链表141.环形链表142.环形链表Ⅱ 160.相交链表 21.合并两个有序链表

Posted 2022-02-09 王嘻嘻-

目录

剑指21.删除链表的倒数第n个节点

剑指22.链表的倒数第k个节点

234.回文链表

141.环形链表

142.环形链表Ⅱ 

160.相交链表

 21.合并两个有序链表​

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剑指21.删除链表的倒数第n个节点

思路解析： 

源代码👇

https://leetcode-cn.com/problems/SLwz0R/submissions/

package leetcode;

public class offer_21 
    //删除链表的倒数第n个节点
    public ListNode removeNthFromEnd(ListNode head, int n) 
        ListNode dummyHead = new ListNode(0);
        ListNode cur = dummyHead;
        dummyHead.next = head;
        ListNode prev = null;
        int i =1;
        while (head!=null)
            if (i>=n)
                prev = cur;
                cur = cur.next;
            
            head = head.next;
            i++;
        
        prev.next = prev.next.next;
        return dummyHead.next;
    

剑指22.链表的倒数第k个节点

思路解析：

 源代码👇

力扣

package leetcode;

public class offer_22 
    //链表中倒数第K个节点
    public ListNode getKthFromEnd(ListNode head, int k) 
        ListNode fast = head,low = head;
        for (int i = 0; i < k; i++) 
            fast = fast.next; //fast先走k个节点
        
        while (fast!=null)
            fast = fast.next;
            low = low.next;
        
        return low; //fast为空，此时low节点即为所求
    

234.回文链表

思路解析：

 源代码👇

https://leetcode-cn.com/problems/palindrome-linked-list/submissions/

package leetcode;
// 回文链表 = 中间节点 + 反转链表
public class Num234 
    public boolean isPalindrome(ListNode head) 
        // 1.找到中间节点
        ListNode middleNode = middleNode(head);
        // 2.反转中间节点后的子链表
        ListNode l2 = reverseList(middleNode);
        // 同时遍历原先链表l1
        while (l2!=null)
            if (head.val!=l2.val)
                return false;
            
            l2 = l2.next;
            head = head.next;
        
        return true;
    
    public ListNode middleNode(ListNode head) 
        //fast一次走两步，low一次走一步
        ListNode fast = head;
        ListNode low = head;
        while (fast!=null&&fast.next!=null)
            fast=fast.next.next;
            low=low.next;
        
        return low;
    
    public ListNode reverseList(ListNode head) 
        if (head==null||head.next==null)
            return head;
        
        ListNode sec = head.next;
        // 反转第二个节点之后的子链表
        ListNode newHead = reverseList(head.next);
        sec.next = head;
        head.next = null;
        return newHead;
    

141.环形链表

思路：采用快慢指针法，如果快指针指向不为空，并且快指针的下一个节点不为空，快指针一次移动两个单位，慢指针一次移动一个单位，若链表有环，快慢指针必会相遇。

源代码👇

https://leetcode-cn.com/problems/palindrome-linked-list/submissions/

package leetcode;
//环形链表
public class Num141 
    public boolean hasCycle(ListNode head) 
        ListNode fast = head,low = head;
        while (fast!=null && fast.next!=null)
            fast = fast.next.next;
            low = low.next;
            if (fast == low)
                return true;
            
        
        return false;
    

142.环形链表Ⅱ 

思路解析： 

 

 源代码👇

https://leetcode-cn.com/problems/palindrome-linked-list/submissions/

package leetcode;
//环形链表Ⅱ  环的入口
public class Num142 
    public ListNode detectCycle(ListNode head) 
        ListNode fast = head,low = head;
        while (fast!=null&&fast.next!=null)
            fast = fast.next.next;
            low = low.next;
            //此时low和fast相遇
            if (fast==low)
                ListNode third = head;
                while (third!=low)
                    third = third.next;
                    low = low.next;
                
                return low;
            
        
        return null;
    

160.相交链表

 思路解析： 

源代码👇

 https://leetcode-cn.com/problems/palindrome-linked-list/submissions/

package leetcode;
//相交链表
public class Num160 
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) 
        ListNode pA = headA;
        ListNode pB = headB;
        // 相交时刚好返回交点, 不相交两个同时走到null
        while (pA!=pB)
            pA = pA == null ? headB : pA.next;
            pB = pB == null ? headA : pB.next;
        
        return pA;
    

 21.合并两个有序链表

思路：采用尾插法，先考虑边界条件，也就是 list1 和 list2 各为空时分别返回 list2 和 list1 ,当两个子链表都不为空，如果 list1 中的值大于 list2 中节点的值，直接在虚拟头节点后拼接 list1 即可。 

源代码👇

https://leetcode-cn.com/problems/palindrome-linked-list/submissions/

package leetcode;
//合并两个有序链表
public class Num21 
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) 
        ListNode dummyHead = new ListNode(101);
        if (list1==null)
            return list2;
        
        if (list2==null)
            return list1;
        
        ListNode last = dummyHead;
        // l1和l2都不为空
        while (list1!=null && list2!=null)
            if (list1.val<= list2.val)
                last.next = list1;
                last = list1;
                list1 = list1.next;
            else
                last.next = list2;
                last = list2;
                list2 = list2.next;
            
        
        // 此时l1或l2为空
        if (list1==null)
            last.next = list2;
        
        if (list2==null)
            last.next = list1;
        
        return dummyHead.next;
    

递归法： 

package leetcode;
//合并两个有序链表
public class Num21 
// 递归法
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) 
        if (list1 == null) 
            return list2;
        
        if (list2 == null) 
            return list1;
        
        if (list1.val <= list2.val) 
            // 说明此时要把list1.next和整个list2交给merge
            // 然后将结果拼接到list1.next
            list1.next = mergeTwoLists(list1.next,list2);
            return list1;
        else 
            list2.next = mergeTwoLists(list1,list2.next);
            return list2;
        
    

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本小节完^_^