基础算法面试题---链表

前言

一般链表的基础题算法都很简单，但却是常见的面试题，因为链表能够考察面试者的编码能力，往往很容易想到解题方式，却写不出来。

下面总结了几道常见的初级题，可以反复练习，提高自己的编码能力。

先准备两个对象，一个单链表，一个双链表

单链表

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;
    }

    @Override
    public String toString() {
     
        return "ListNode{" +
                "val=" + val +
                ", next=" + next +
                '}';
    }
}

双链表

public class DoubleListNode<T> {
     
    public T data;
    public DoubleListNode<T> last;
    public DoubleListNode<T> next;

    public DoubleListNode(T data) {
     
        this.data = data;
    }

    @Override
    public String toString() {
     
        return "DoubleListNode{" +
                "data=" + data +
                ", next=" + next +
                '}';
    }
}

1、单链表反转

单链表反转，经典且基础的链表题。

举例：
原链表：1-2-3-4-5
反转后：5-4-3-2-1

public class Code_01 {
     
    public static void main(String[] args) {
     
        ListNode n = new ListNode();
        ListNode head = n;
        for (int i = 1; i <= 5; i++) {
     
            n.next = new ListNode(i);
            n = n.next;
        }
        System.out.println("原链表：" + head);
        ListNode listNode = reverseList(head);
        System.out.println("反转后：" + listNode);
    }

    private static ListNode reverseList(ListNode head) {
     
        ListNode pre = null;
        ListNode next;
        while (head != null) {
     
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}

2、双链表反转

public class Code_02 {
     
    public static void main(String[] args) {
     
        DoubleListNode n1 = new DoubleListNode(1);
        DoubleListNode n2 = new DoubleListNode(2);
        n1.next = n2;
        n2.last = n1;
        DoubleListNode n3 = new DoubleListNode(3);
        n2.next = n3;
        n3.last = n2;
        System.out.println("原链表：" + n1);
        DoubleListNode listNode = reverseDoubleList(n1);
        System.out.println("反转后：" + listNode);
    }

    private static DoubleListNode reverseDoubleList(DoubleListNode head) {
     
        DoubleListNode pre = null;
        DoubleListNode next;
        while (head != null) {
     
            next = head.next;
            head.next = pre;
            head.last = next;
            pre = head;
            head = next;
        }
        return pre;
    }
}

3、删除链表中的某类节点

举例：
原链表：1-2-3-4-5
删除value=2的节点
结果：1-3-4-5

原链表：1-2-3-3-3-4-5
删除value=3的节点
结果：1-2-4-5

原链表：1-1-2-3-4-5
删除value=1的节点
结果：2-3-4-5

public class Code_03 {
     
    public static void main(String[] args) {
     
        ListNode n = new ListNode();
        ListNode head = n;
        for (int i = 1; i <= 5; i++) {
     
            n.next = new ListNode(i);
            n = n.next;
        }
        System.out.println("原链表：" + head);
        ListNode node = deleteNode(head, 3);
        System.out.println("删除后：" + node);
    }

    private static ListNode deleteNode(ListNode head, int value) {
     
        //处理head本身就是要删除的节点
        while (head != null) {
     
            if (head.val != value) {
     
                break;
            }
            head = head.next;
        }
        //始终记录前一个节点，和当前节点的指针，如果当前节点就是要删除的节点时，则让前一个节点指向当前节点的下一个节点，即完成了删除
        ListNode cur = head;
        ListNode pre = null;
        while (cur != null) {
     
            if (cur.val == value) {
     
                pre.next = cur.next;
            } else {
     
                pre = cur;
            }
            cur = cur.next;
        }
        return head;
    }
}

4、删除重复元素1

给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。

举例：
原链表: 1->1->2->3->3
删除后: 1->2->3

public class Code_05 {
     
    public static void main(String[] args) {
     
        Code_05 c = new Code_05();
        ListNode n = new ListNode();
        ListNode head = n;
        for (int i = 1; i <= 5; i++) {
     
            n.next = new ListNode(i);
            n = n.next;
            if (i % 2 == 0) {
     
                n.next = new ListNode(i);
                n = n.next;
            }
        }
        System.out.println("原链表：" + head);
        System.out.println("删除后：" + c.deleteDuplicates(head));
    }

    /**
     * 通过不断移动cur，判断当前cur的值与cur.next的值是否相等，如果相等，则只改变cur.next，并让其指向下一个节点，就等于跳过了cur.next的节点
     * 如果不相等，则移动cur节点位置到cur.next上。
     * @param head
     * @return
     */
    public ListNode deleteDuplicates(ListNode head) {
     
        ListNode cur = head;
        while (cur != null && cur.next != null) {
     
            if (cur.val != cur.next.val) {
     
                cur = cur.next;
            } else {
     
                cur.next = cur.next.next;
            }
        }
        return head;
    }
}

5、删除重复元素2

给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中没有重复出现的数字。

举例：
原链表: 1->2->3->3->4->4->5
删除后: 1->2->5

这是在前一道题目上的延伸。

解法一：

结合前两题的方式，可以拆分处理，先用第4题的方式删除重复的数字并记录下来，再第3题的方式删除指定数字。

public class Code_06 {
     

    public static void main(String[] args) {
     
        Code_06 c = new Code_06();
        ListNode n = new ListNode();
        ListNode head = n;
        for (int i = 1; i <= 5; i++) {
     
            n.next = new ListNode(i);
            n = n.next;
            if (i % 2 == 0) {
     
                n.next = new ListNode(i);
                n = n.next;
            }
        }
        System.out.println(head);
        System.out.println(c.deleteDuplicates(head));
    }

    public ListNode deleteDuplicates(ListNode head) {
     
        ListNode cur = head;
        Set<Integer> dupSet = new HashSet<>();
        while (cur != null && cur.next != null) {
     
            if (cur.val != cur.next.val) {
     
                cur = cur.next;
            } else {
     
                dupSet.add(cur.val);
                cur.next = cur.next.next;
            }
        }
        for (Integer val : dupSet) {
     
            head = deleteNode(head, val);
        }
        return head;
    }

    private ListNode deleteNode(ListNode head, int val) {
     
        while (head != null) {
     
            if (head.val != val) {
     
                break;
            }
            head = head.next;
        }
        ListNode pre = null;
        ListNode cur = head;
        while (cur != null) {
     
            if (cur.val == val) {
     
                pre.next = cur.next;
            } else {
     
                pre = cur;
            }
            cur = cur.next;
        }
        return head;
    }

}

当然第一种解法只能当做是编码的练习，出题者肯定不是希望你用这种方式处理。

解法二：

哑节点+双指针

public class Code_06_01 {
     
    public static void main(String[] args) {
     
        Code_06_01 c = new Code_06_01();
        ListNode n = new ListNode();
        ListNode head = n;
        for (int i = 1; i <= 5; i++) {
     
            n.next = new ListNode(i);
            n = n.next;
            if (i % 2 == 0) {
     
                n.next = new ListNode(i);
                n = n.next;
            }
        }
        System.out.println(head);
        System.out.println(c.deleteDuplicates(head));
    }

    /**
     * 哑节点+双指针
     * 

     * 构建一个哑节点，让其next指向头位置。
     * 再利用双指针，n1,n2,初始都指向head位置，如果n1.next.val==n2.next.val，则让n2向前移动一位，否则n1，n2一起向前移动一位
     *
     * @param head
     * @return
     */
    public ListNode deleteDuplicates(ListNode head) {
     
        ListNode dummy = new ListNode();
        dummy.next = head;
        ListNode n1 = dummy;
        ListNode n2 = head;
        while (n2 != null && n2.next != null) {
     
            if (n1.next.val != n2.next.val) {
     
                n1 = n1.next;
            } else {
     
                //n2一直移动，直到不等于n1为止
                while (n2 != null && n2.next != null && n1.next.val == n2.next.val) {
     
                    n2 = n2.next;
                }
                n1.next = n2.next;
            }
            n2 = n2.next;
        }
        return dummy.next;
    }
}

6、合并有序链表

将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

原链表：l1 = [1,2,4]，l2 = [1,3,4]
合并后：[1,1,2,3,4,4]

public class Code_04 {
     

    public static void main(String[] args) {
     
        ListNode n = new ListNode(1);
        ListNode l1 = n;
        for (int i = 3; i <= 5; i = i + 2) {
     
            n.next = new ListNode(i);
            n = n.next;
        }
        ListNode n2 = new ListNode(2);
        ListNode l2 = n2;
        for (int i = 4; i <= 6; i = i + 2) {
     
            n2.next = new ListNode(i);
            n2 = n2.next;
        }
        Code_04 c = new Code_04();

        System.out.println("原链表1：" + l1);
        System.out.println("原链表2：" + l2);

        System.out.println("合并后：" + c.mergeTwoLists(l1, l2));

    }

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
     
        if (l1 == null) {
     
            return l2;
        }
        if (l2 == null) {
     
            return l1;
        }
        ListNode mergeNode = new ListNode();
        ListNode pre = mergeNode;
        //比较两个链表当前的值，值小的链表就把引用赋给mergeNode，并向后移动一位重新赋值给自己，同时pre指向值小的那个节点
        while (l1 != null && l2 != null) {
     
            if (l1.val <= l2.val) {
     
                pre.next = l1;
                l1 = l1.next;
            } else {
     
                pre.next = l2;
                l2 = l2.next;
            }
            pre = pre.next;
        }
        pre.next = l1 == null ? l2 : l1;

        return mergeNode.next;
    }
}

7、双向链表实现栈、队列

队列：先进先出

栈：先进后出

分别用双向链表实现，从头进，从头出，从尾进，从尾出即可模拟出队列和栈的数据结构。

public class NodeUtils<T> {
     

    DoubleListNode<T> head;
    DoubleListNode<T> tail;

    /**
     * 从头进
     * @param data
     */
    public void addHead(T data) {
     
        DoubleListNode<T> node = new DoubleListNode(data);
        if (head == null) {
     
            head = node;
            tail = node;
        } else {
     
            node.next = head;
            head.last = node;
            head = node;
        }
    }

    /**
     * 从尾进
     * @param data
     */
    public void addTail(T data) {
     
        DoubleListNode<T> node = new DoubleListNode(data);
        if (head == null) {
     
            head = node;
        } else {
     
            tail.next = node;
            node.last = tail;
        }
        tail = node;
    }

    /**
     * 从头出
     * @return
     */
    public T popHead() {
     
        if (head == null) {
     
            return null;
        }
        DoubleListNode<T> h = head;
        if (head == tail) {
     
            head = null;
            tail = null;
        } else {
     
            head = head.next;
            head.last = null;
        }
        return h.data;
    }

    /**
     * 从尾出
     * @return
     */
    public T popTail() {
     
        if (tail == null) {
     
            return null;
        }
        DoubleListNode<T> t = tail;
        if (head == tail) {
     
            head = null;
            tail = null;
        } else {
     
            tail = tail.last;
            tail.next = null;
        }
        return t.data;

    }

}

队列

/**
 * 队列（先进先出）：从链表头进，从链表尾出
 * @param 
 */
public class ListNodeQueue<T> {
     

    NodeUtils<T> nodeUtils = new NodeUtils();

    public void push(T data) {
     
        nodeUtils.addHead(data);
    }

    public T pop() {
     
        return nodeUtils.popTail();
    }

}

栈

/**
 * 队列（先进先出）：从链表头进，从链表尾出
 *
 * @param 
 */
public class ListNodeQueue<T> {
     

    NodeUtils<T> nodeUtils = new NodeUtils();

    public void push(T data) {
     
        nodeUtils.addHead(data);
    }

    public T pop() {
     
        return nodeUtils.popTail();
    }

}

验证

public class Code_07 {
     
    public static void main(String[] args) {
     
        ListNodeStack stack = new ListNodeStack();
        stack.push(1);
        stack.push(2);
        stack.push(3);
        System.out.println("栈：压进去1,2,3");
        System.out.println("弹出" + stack.pop() + "," + stack.pop() + "," + stack.pop());

        ListNodeQueue queue = new ListNodeQueue();
        queue.push(1);
        queue.push(2);
        queue.push(3);
        System.out.println("队列：压进去1,2,3");
        System.out.println("弹出" + queue.pop() + "," + queue.pop() + "," + queue.pop());
    }
}