链表

Posted 2021-03-15 wwj99

哈希表的简单介绍

1） 哈希表在使用层面上可以理解为一种集合结构

2） 如果只有key，没有伴随数据value,可以使用HashSet结构（C++中叫UnOrderedSet）

3） 如果既有key，又有伴随数据value，可以使用HashMap结构（C++中叫UnOrderedMap）

4） 有无伴随数据，是HashMap和HashSet唯一的区别，底层的实际结构是一回事

5） 使用哈希表增（put）、删（remove）、改（put）和查（get ）的操作,可以认为时间复杂度为O(1)，但是常数时间比较大

6） 放入哈希表的东西，如果是基础类型，内部按值传递，内存占用就是这个东西的大小

7） 放入哈希表的东西，如果不是基础类型，内部按引用传递，内存占用是这个东西内存地址的大小

有序表的简单介绍

1） 有序表在使用层面上可以理解为一种集合结构

2） 如果只有key，没有伴随数据value，可以使用TreeSet结构（C++中叫OrderedSet）

3） 如果既有key，又有伴随数据value，可以使用TreeMap结构（C++中叫OrderedMap）

4） 有无伴随数据，是TreeSet和TreeMap唯一的区别，底层的实际结构是一回事

5） 有序表和哈希表的区别是，有序表把key按照顺序组织起来，而哈希表完全不组织

5） 红黑树、AVL树、size-balance-tree和跳表等都属于有序表结构，只是底层具体实现 不同

6） 放入有序表的东西，如果是基础类型，内部按值传递，内存占用就是这个东西的大小

7） 放入有序表的东西，如果不是基础类型，必须提供比较器，内部按引用传递，内存占用是这个东西内存地址的大小

8） 不管是什么底层具体实现，只要是有序表，都有以下固定的基本功能和固定的时间复杂度

有序表的固定操作

1) void put (K key, V va l ue):将一个(key, va l ue)记录加入到表中，或者将key的记录更新成value

2) V get (K key):根据给定的key，查询value并返回。

3) void remove (K key):移除key的记录。

4) boolean containsKey (K key):询问是否有关于key的记录。

5) K firstKey ():返回所有键值的排序结果中，最左(最小)的那个。

6) K lastKey()：返回所有键值的排序结果中，最右(最大)的那个。

7) K floorKey (K key):如果表中存入过key，返回key；否则返回所有键值的排序结果，key的前一个。

8) K ceilingKey(K key):如果表中存入过key，返回key；否则返回所有键值的排序结果中, key的后一个。

以上所有操作时间复杂度都是0(IogN)

import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.TreeMap;
import java.util.TreeSet;

public class HashAndTree {

    public static class Node {
        public int value;
        public Node next;

        public Node(int val) {
            value = val;
        }
    }

    public static class NodeComparator implements Comparator<Node> {

        @Override
        public int compare(Node o1, Node o2) {
            return o1.value - o2.value;
        }

    }

    public static void main(String[] args) {
        Node nodeA = null;
        Node nodeB = null;
        Node nodeC = null;

        // hashSet1的key是基础类型->int类型
        HashSet<Integer> hashSet1 = new HashSet<>();
        hashSet1.add(3);
        System.out.println(hashSet1.contains(3));
        hashSet1.remove(3);
        System.out.println(hashSet1.contains(3));

        // hashSet2的key是非基础类型->Node类型
        nodeA = new Node(1);
        nodeB = new Node(1);
        HashSet<Node> hashSet2 = new HashSet<>();
        hashSet2.add(nodeA);
        System.out.println(hashSet2.contains(nodeA));
        System.out.println(hashSet2.contains(nodeB));
        hashSet2.remove(nodeA);
        System.out.println(hashSet2.contains(nodeA));

        // hashMap1的key是基础类型->String类型
        HashMap<String, Integer> hashMap1 = new HashMap<>();
        String str1 = "key";
        String str2 = "key";
        hashMap1.put(str1, 1);
        System.out.println(hashMap1.containsKey(str1));
        System.out.println(hashMap1.containsKey(str2));
        System.out.println(hashMap1.get(str1));
        System.out.println(hashMap1.get(str2));

        hashMap1.put(str2, 2);
        System.out.println(hashMap1.containsKey(str1));
        System.out.println(hashMap1.containsKey(str2));
        System.out.println(hashMap1.get(str1));
        System.out.println(hashMap1.get(str2));

        hashMap1.remove(str1);
        System.out.println(hashMap1.containsKey(str1));
        System.out.println(hashMap1.containsKey(str2));

        // hashMap2的key是非基础类型->Node类型
        nodeA = new Node(1);
        nodeB = new Node(1);
        HashMap<Node, String> hashMap2 = new HashMap<>();
        hashMap2.put(nodeA, "A节点");
        System.out.println(hashMap2.containsKey(nodeA));
        System.out.println(hashMap2.containsKey(nodeB));
        System.out.println(hashMap2.get(nodeA));
        System.out.println(hashMap2.get(nodeB));
        hashMap2.put(nodeB, "B节点");
        System.out.println(hashMap2.containsKey(nodeA));
        System.out.println(hashMap2.containsKey(nodeB));
        System.out.println(hashMap2.get(nodeA));
        System.out.println(hashMap2.get(nodeB));

        // treeSet的key是非基础类型->Node类型
        nodeA = new Node(5);
        nodeB = new Node(3);
        nodeC = new Node(7);

        treeSet = new TreeSet<>(new NodeComparator());
        // 要提供Node类型的比较器
        try {
            treeSet.add(nodeA);
            treeSet.add(nodeB);
            treeSet.add(nodeC);
            System.out.println("这次节点都加入了");
        } catch (Exception e) {
            System.out.println(e.getMessage());
        }

        // 展示有序表常用操作
        TreeMap<Integer, String> treeMap1 = new TreeMap<>();
        treeMap1.put(7, "我是7");
        treeMap1.put(5, "我是5");
        treeMap1.put(4, "我是4");
        treeMap1.put(3, "我是3");
        treeMap1.put(9, "我是9");
        treeMap1.put(2, "我是2");
        System.out.println(treeMap1.containsKey(5));
        System.out.println(treeMap1.get(5));
        System.out.println(treeMap1.firstKey() + ", 我最小");
        System.out.println(treeMap1.lastKey() + ", 我最大");
        System.out.println(treeMap1.floorKey(8) + ", 在表中所有<=8的数中，我离8最近");
        System.out.println(treeMap1.ceilingKey(8) + ", 在表中所有>=8的数中，我离8最近");
        System.out.println(treeMap1.floorKey(7) + ", 在表中所有<=7的数中，我离7最近");
        System.out.println(treeMap1.ceilingKey(7) + ", 在表中所有>=7的数中，我离7最近");
        treeMap1.remove(5);
        System.out.println(treeMap1.get(5) + ", 删了就没有了哦");
    }
}

单链表的节点结构

Class Node< v > {

? V value;

? Node next; }

由以上结构的节点依次连接起来所形成的链叫单链表结构。

双链表的节点结构

Class Node< V >{

? V value;

? Node next;

? Node last; }

由以上结构的节点依次连接起来所形成的链叫双链表结构。

单链表和双链表结构只需要给定一个头部节点head，就可以找到剩下的所有的节点。

反转单向和双向链表

【题目】分别实现反转单向链表和反转双向链表的函数

【要求】如果链表长度为N,时间复杂度要求为0(N),额外空间复杂度要求为 0(1)

public class ReverseList {

    public static class Node {
        public int value;
        public Node next;

        public Node(int data) {
            this.value = data;
        }
    }

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

    public static class DoubleNode {
        public int value;
        public DoubleNode last;
        public DoubleNode next;

        public DoubleNode(int data) {
            this.value = data;
        }
    }

    public static DoubleNode reverseList(DoubleNode head) {
        DoubleNode pre = null;
        DoubleNode next = null;
        while (head != null) {
            next = head.next;
            head.next = pre;
            head.last = next;
            pre = head;
            head = next;
        }
        return pre;
    }

    public static void printLinkedList(Node head) {
        System.out.print("Linked List: ");
        while (head != null) {
            System.out.print(head.value + " ");
            head = head.next;
        }
        System.out.println();
    }

    public static void printDoubleLinkedList(DoubleNode head) {
        System.out.print("Double Linked List: ");
        DoubleNode end = null;
        while (head != null) {
            System.out.print(head.value + " ");
            end = head;
            head = head.next;
        }
        System.out.print("| ");
        while (end != null) {
            System.out.print(end.value + " ");
            end = end.last;
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Node head1 = new Node(1);
        head1.next = new Node(2);
        head1.next.next = new Node(3);
        printLinkedList(head1);
        head1 = reverseList(head1);
        printLinkedList(head1);

        DoubleNode head2 = new DoubleNode(1);
        head2.next = new DoubleNode(2);
        head2.next.last = head2;
        head2.next.next = new DoubleNode(3);
        head2.next.next.last = head2.next;
        head2.next.next.next = new DoubleNode(4);
        head2.next.next.next.last = head2.next.next;
        printDoubleLinkedList(head2);
        printDoubleLinkedList(reverseList(head2));
    }
}

打印两个有序链表的公共部分

【题目】给定两个有序链表的头指针head 1和head2,打印两个链表的公共部分。

【要求】如果两个链表的长度之和为N,时间复杂度要求为0(N),额外空间复 杂度要求为0(1)

public class PrintCommonPart {

    public static class Node {
        public int value;
        public Node next;
        public Node(int data) {
            this.value = data;
        }
    }

    public static void printCommonPart(Node head1, Node head2) {
        System.out.print("Common Part: ");
        while (head1 != null && head2 != null) {
            if (head1.value < head2.value) {
                head1 = head1.next;
            } else if (head1.value > head2.value) {
                head2 = head2.next;
            } else {
                System.out.print(head1.value + " ");
                head1 = head1.next;
                head2 = head2.next;
            }
        }
        System.out.println();
    }

    public static void printLinkedList(Node node) {
        System.out.print("Linked List: ");
        while (node != null) {
            System.out.print(node.value + " ");
            node = node.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Node node1 = new Node(2);
        node1.next = new Node(3);
        node1.next.next = new Node(5);
        node1.next.next.next = new Node(6);

        Node node2 = new Node(1);
        node2.next = new Node(2);
        node2.next.next = new Node(5);
        node2.next.next.next = new Node(7);
        node2.next.next.next.next = new Node(8);

        printLinkedList(node1);
        printLinkedList(node2);
        printCommonPart(node1, node2);
    }
}

面试时链表解题的方法论

1） 对于笔试，不用太在乎空间复杂度，一切为了时间复杂度

2） 对于面试，时间复杂度依然放在第一位，但是一定要找到空间最省的方法

重要技巧：

1） 额外数据结构记录（哈希表等）

2） 快慢指针(熟悉code)

判断一个链表是否为回文结构

【题目】给定一个单链表的头节点head，请判断该链表是否为回文结构。

【例子】1->2->1，返回true； 1->2->2->1，返回true； 15->6->15，返回true； 1->2->3，返回false。

【例子】如果链表长度为N，时间复杂度达到0(N)，额外空间复杂度达到0(1)。

import java.util.Stack;

public class Code04_IsPalindromeList {

    public static class Node {
        public int value;
        public Node next;

        public Node(int data) {
            this.value = data;
        }
    }

    // need n extra space
    public static boolean isPalindrome1(Node head) {
        Stack<Node> stack = new Stack<Node>();
        Node cur = head;
        while (cur != null) {
            stack.push(cur);
            cur = cur.next;
        }
        while (head != null) {
            if (head.value != stack.pop().value) {
                return false;
            }
            head = head.next;
        }
        return true;
    }

    // need n/2 extra space
    public static boolean isPalindrome2(Node head) {
        if (head == null || head.next == null) {
            return true;
        }
        Node right = head.next;
        Node cur = head;
        while (cur.next != null && cur.next.next != null) {
            right = right.next;
            cur = cur.next.next;
        }
        Stack<Node> stack = new Stack<Node>();
        while (right != null) {
            stack.push(right);
            right = right.next;
        }
        while (!stack.isEmpty()) {
            if (head.value != stack.pop().value) {
                return false;
            }
            head = head.next;
        }
        return true;
    }

    // need O(1) extra space
    public static boolean isPalindrome3(Node head) {
        if (head == null || head.next == null) {
            return true;
        }
        Node n1 = head;
        Node n2 = head;
        while (n2.next != null && n2.next.next != null) { // find mid node
            n1 = n1.next; // n1 -> mid
            n2 = n2.next.next; // n2 -> end
        }
        n2 = n1.next; // n2 -> right part first node
        n1.next = null; // mid.next -> null
        Node n3 = null;
        while (n2 != null) { // right part convert
            n3 = n2.next; // n3 -> save next node
            n2.next = n1; // next of right node convert
            n1 = n2; // n1 move
            n2 = n3; // n2 move
        }
        n3 = n1; // n3 -> save last node
        n2 = head;// n2 -> left first node
        boolean res = true;
        while (n1 != null && n2 != null) { // check palindrome
            if (n1.value != n2.value) {
                res = false;
                break;
            }
            n1 = n1.next; // left to mid
            n2 = n2.next; // right to mid
        }
        n1 = n3.next;
        n3.next = null;
        while (n1 != null) { // recover list
            n2 = n1.next;
            n1.next = n3;
            n3 = n1;
            n1 = n2;
        }
        return res;
    }

    public static void (Node node) {
        System.out.print("Linked List: ");
        while (node != null) {
            System.out.print(node.value + " ");
            node = node.next;
        }
        System.out.println();
    }
    public static void main(String[] args) {

        Node head = null;
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

        head = new Node(1);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

        head = new Node(1);
        head.next = new Node(2);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

        head = new Node(1);
        head.next = new Node(1);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(1);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(1);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(2);
        head.next.next.next = new Node(1);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(2);
        head.next.next.next.next = new Node(1);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);
    }
}

将单向链表按某值划分成左边小、中间相等、右边大的形式

【题目】给定一个单链表的头节点head，节点的值类型是整型，再给定一个整 数pivoto实现一个调整链表的函数，将链表调整为左部分都是值小于pivot的 节点，中间部分都是值等于pivot的节点，右部分都是值大于pivot的节点。(各种各样的情况，讨论好边界)

【进阶】在实现原问题功能的基础上增加如下的要求

【要求】调整后所有小于pivot的节点之间的相对顺序和调整前一样，调整后所有等于pivot的节点之间的相对顺序和调整前一样，调整后所有大于pivot的节点之间的相对顺序和调整前一样

【要求】时间复杂度请达到0(N),额外空间复杂度请达到0(1)。

public class SmallerEqualBigger {

    public static class Node {
        public int value;
        public Node next;

        public Node(int data) {
            this.value = data;
        }
    }

    public static Node listPartition1(Node head, int pivot) {
        if (head == null) {
            return head;
        }
        Node cur = head;
        int i = 0;
        while (cur != null) {
            i++;
            cur = cur.next;
        }
        Node[] nodeArr = new Node[i];
        i = 0;
        cur = head;
        for (i = 0; i != nodeArr.length; i++) {
            nodeArr[i] = cur;
            cur = cur.next;
        }
        arrPartition(nodeArr, pivot);
        for (i = 1; i != nodeArr.length; i++) {
            nodeArr[i - 1].next = nodeArr[i];
        }
        nodeArr[i - 1].next = null;
        return nodeArr[0];
    }

    public static void arrPartition(Node[] nodeArr, int pivot) {
        int small = -1;
        int big = nodeArr.length;
        int index = 0;
        while (index != big) {
            if (nodeArr[index].value < pivot) {
                swap(nodeArr, ++small, index++);
            } else if (nodeArr[index].value == pivot) {
                index++;
            } else {
                swap(nodeArr, --big, index);
            }
        }
    }

    public static void swap(Node[] nodeArr, int a, int b) {
        Node tmp = nodeArr[a];
        nodeArr[a] = nodeArr[b];
        nodeArr[b] = tmp;
    }

    public static Node listPartition2(Node head, int pivot) {
        Node sH = null; // small head
        Node sT = null; // small tail
        Node eH = null; // equal head
        Node eT = null; // equal tail
        Node bH = null; // big head
        Node bT = null; // big tail
        Node next = null; // save next node
        // every node distributed to three lists
        while (head != null) {
            next = head.next;
            head.next = null;
            if (head.value < pivot) {
                if (sH == null) {
                    sH = head;
                    sT = head;
                } else {
                    sT.next = head;
                    sT = head;
                }
            } else if (head.value == pivot) {
                if (eH == null) {
                    eH = head;
                    eT = head;
                } else {
                    eT.next = head;
                    eT = head;
                }
            } else {
                if (bH == null) {
                    bH = head;
                    bT = head;
                } else {
                    bT.next = head;
                    bT = head;
                }
            }
            head = next;
        }
        // small and equal reconnect
        if (sT != null) {  //如果有小于区域
            sT.next = eH;
            eT = eT == null ? sT : eT;   //下一步，谁去连大于区域的头，谁就变成eT
        }
        // all reconnect
        if (eT != null) {  // 如果小于区域和等于区域，不是都没有
            eT.next = bH;
        }
        return sH != null ? sH : eH != null ? eH : bH;
    }

    public static void printLinkedList(Node node) {
        System.out.print("Linked List: ");
        while (node != null) {
            System.out.print(node.value + " ");
            node = node.next;
        }
        System.out.println();
    }
    
    public static void main(String[] args) {
        Node head1 = new Node(7);
        head1.next = new Node(9);
        head1.next.next = new Node(1);
        head1.next.next.next = new Node(8);
        head1.next.next.next.next = new Node(5);
        head1.next.next.next.next.next = new Node(2);
        head1.next.next.next.next.next.next = new Node(5);
        printLinkedList(head1);
        // head1 = listPartition1(head1, 4);
        head1 = listPartition2(head1, 5);
        printLinkedList(head1);

    }
}

复制含有随机指针节点的链表

【题目】一种特殊的单链表节点类描述如下

class Node {

? int value;

? Node next;

? Node rand;

? Node (int val) {

? value = va l ;

? }

}

rand指针是单链表节点结构中新增的指针，rand可能指向链表中的任意一个节 点，也可能指向null。给定一个由Node节点类型组成的无环单链表的头节点 head，请实现一个函数完成这个链表的复制，并返回复制的新链表的头节点。

【要求】时间复杂度0(N),额外空间复杂度0(1)

import java.util.HashMap;

public class CopyListWithRandom {

    public static class Node {
        public int value;
        public Node next;
        public Node rand;

        public Node(int data) {
            this.value = data;
        }
    }

    public static Node copyListWithRand1(Node head) {
        HashMap<Node, Node> map = new HashMap<Node, Node>();
        Node cur = head;
        while (cur != null) {
            map.put(cur, new Node(cur.value));
            cur = cur.next;
        }
        cur = head;
        while (cur != null) {
            map.get(cur).next = map.get(cur.next);
            map.get(cur).rand = map.get(cur.rand);
            cur = cur.next;
        }
        return map.get(head);
    }

    public static Node copyListWithRand2(Node head) {
        if (head == null) {
            return null;
        }
        Node cur = head;
        Node next = null;
        // copy node and link to every node
        //1 -> 2
        //1 -> 1' -> 2
        while (cur != null) {
            next = cur.next;
            cur.next = new Node(cur.value);
            cur.next.next = next;
            cur = next;
        }
        cur = head;
        Node curCopy = null;
        // set copy node rand
        // 1 -> 1' -> 2 -> 2'
        while (cur != null) {
            next = cur.next.next;
            curCopy = cur.next;
            curCopy.rand = cur.rand != null ? cur.rand.next : null;
            cur = next;
        }
        Node res = head.next;
        cur = head;
        // split
        while (cur != null) {
            next = cur.next.next;
            curCopy = cur.next;
            cur.next = next;
            curCopy.next = next != null ? next.next : null;
            cur = next;
        }
        return res;
    }

    public static void printRandLinkedList(Node head) {
        Node cur = head;
        System.out.print("order: ");
        while (cur != null) {
            System.out.print(cur.value + " ");
            cur = cur.next;
        }
        System.out.println();
        cur = head;
        System.out.print("rand:  ");
        while (cur != null) {
            System.out.print(cur.rand == null ? "- " : cur.rand.value + " ");
            cur = cur.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Node head = null;
        Node res1 = null;
        Node res2 = null;
        printRandLinkedList(head);
        res1 = copyListWithRand1(head);
        printRandLinkedList(res1);
        res2 = copyListWithRand2(head);
        printRandLinkedList(res2);
        printRandLinkedList(head);

        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);
        head.next.next.next.next.next = new Node(6);

        head.rand = head.next.next.next.next.next; // 1 -> 6
        head.next.rand = head.next.next.next.next.next; // 2 -> 6
        head.next.next.rand = head.next.next.next.next; // 3 -> 5
        head.next.next.next.rand = head.next.next; // 4 -> 3
        head.next.next.next.next.rand = null; // 5 -> null
        head.next.next.next.next.next.rand = head.next.next.next; // 6 -> 4

        printRandLinkedList(head);
        res1 = copyListWithRand1(head);
        printRandLinkedList(res1);
        res2 = copyListWithRand2(head);
        printRandLinkedList(res2);
        printRandLinkedList(head);
    }
}

两个单链表相交的一系列问题

【题目】给定两个可能有环也可能无环的单链表，头节点head 1和head2。请实 现一个函数，如果两个链表相交，请返回相交的第一个节点。如果不相交，返 回 nu l l

【要求】如果两个链表长度之和为N,时间复杂度请达到0(N),额外空间复杂度 请达到0(1)。