Java数据结构你必须要掌握的链表面试经典例题（附超详细图解和代码）

Posted 2021-12-08 /少司命

目录

一，写在前面

二，链表经典例题

1，反转一个单链表

2，给定一个带有头结点 head 的非空单链表，返回链表的中间结点

3，输入一个链表，输出该链表中倒数第k个结点

4，删除链表中的多个重复值

5，链表的回文结构

6，合并两个链表

7，输入两个链表，找出它们的第一个公共结点。

8，判断一个链表是否有环

9，求有环链表的环第一个结点

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一，写在前面

链表几乎可以说是数据结构的重中之重，链表也是大厂面试必考的知识点，要想学好链表，最重要的就是画图解决问题，如果你认为此博客写的不错的话，求点赞，求收藏，求评论，你的三连是我进步最大的动力，废话不多说，让我们学起来吧！！！

二，链表经典例题

1，反转一个单链表

 

public ListNode reverseList() {
        if(this.head == null) {
            return null;
        }
        ListNode cur = this.head;
        ListNode prev = null;

        while (cur != null) {
            ListNode curNext = cur.next;
            cur.next = prev;
            prev = cur;
            cur = curNext;
        }
        return prev;
    }

 2，给定一个带有头结点 head 的非空单链表，返回链表的中间结点

public ListNode middleNode() {
        if(head == null) {
            return null;
        }
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            if(fast == null) {
                return slow;
            }
            slow = slow.next;
        }
        return slow;
    }

 3，输入一个链表，输出该链表中倒数第k个结点

 public ListNode findKthToTail(int k) {
        if(k <= 0 || head == null) {
            return null;
        }
        ListNode fast = head;
        ListNode slow = head;
        while (k-1 != 0) {
            fast = fast.next;
            if(fast == null) {
                return null;
            }
            k--;
        }
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }

4，删除链表中的多个重复值

//删除所有值为key的节点
    public ListNode removeAllKey(int key){
        if(this.head == null) return null;
        ListNode prev = this.head;
        ListNode cur = this.head.next;
        while (cur != null) {
            if(cur.val == key) {
                prev.next = cur.next;
                cur = cur.next;
            }else {
                prev = cur;
                cur = cur.next;
            }
        }
        //最后处理头
        if(this.head.val == key) {
            this.head = this.head.next;
        }
        return this.head;
    }

5，链表的回文结构

 public boolean chkPalindrome(ListNode A) {
        // write code here
        if(head == null) return true;
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        //slow走到了中间位置-》反转

        ListNode cur = slow.next;
        while(cur != null) {
            ListNode curNext = cur.next;
            cur.next = slow;
            slow = cur;
            cur = curNext;
        }
        //反转完成
        while(head != slow) {
            if(head.val != slow.val) {
                return false;
            }
            if(head.next == slow) {
                return true;
            }
            head = head.next;
            slow = slow.next;
        }
        return true;
    }

6，合并两个链表

public static ListNode mergeTwoLists(ListNode headA, ListNode headB) {
        ListNode newHead = new ListNode(-1);
        ListNode tmp = newHead;
        while (headA != null && headB != null) {
            if(headA.val < headB.val) {
                tmp.next = headA;
                headA = headA.next;
                tmp = tmp.next;
            }else {
                tmp.next = headB;
                headB = headB.next;
                tmp = tmp.next;
            }
        }
        if(headA != null) {
            tmp.next = headA;
        }
        if(headB != null) {
            tmp.next = headB;
        }
        return newHead.next;
    }

7，输入两个链表，找出它们的第一个公共结点。

 public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if(headA == null || headB == null) {
            return null;
        }

        ListNode pl = headA;
        ListNode ps = headB;
        int lenA = 0;
        int lenB = 0;
        while (pl != null) {
            lenA++;
            pl = pl.next;
        }
        //pl==null
        pl = headA;
        while (ps != null) {
            lenB++;
            ps = ps.next;
        }
        //ps==null
        ps = headB;
        int len = lenA-lenB;//差值步
        if(len < 0) {
            pl = headB;
            ps = headA;
            len = lenB-lenA;
        }
        //1、pl永远指向了最长的链表   ps 永远指向了最短的链表  2、求到了差值len步

        //pl走差值len步
        while (len != 0) {
            pl = pl.next;
            len--;
        }
        //同时走 直到相遇
        while (pl != ps) {
            pl = pl.next;
            ps = ps.next;
        }
        return pl;
    }

8，判断一个链表是否有环

 public boolean hasCycle() {
        if(head == null) return false;
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if(fast == slow) {
                return true;
            }
        }
        return false;
    }

 9，求有环链表的环第一个结点

 public ListNode detectCycle(ListNode head) {
        if(head == null) return null;
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if(fast == slow) {
                break;
            }
        }
        if(fast == null || fast.next == null) {
            return null;
        }
        fast = head;
        while (fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }
        return fast;
    }