链表结构

• 单链表的节点结构  由以下结构的节点依次连接起来所形成的链叫单链表结构

Clas Node<V>{ V value; Node next;
} 

• 双链表的节点结构 由以下结构的节点依次连接起来所形成的链叫双链表结构

Clas Node<V>{ V value; Node next; Node last;
} 

• 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O(N)，额外空间复杂度要求为 O(1)

package com.example.algorithm.demo.class04;public class Code02_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head1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O(N)，额外空间复杂度要求为O(1)
• [思路] head1指向第一个有序链表，head2指向第二个有序链表，比较元素的大小，如果相等打印元素的数值。否则，指针指向的元素的数值小的率先移动，比较大小，元素数值小的移动

package com.example.algorithm.demo.class04;public class Code03_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.println("Common Part: ");while (head1 != null && head2 != null){if (head1.value < head2.value){head1 = head1.next;} else if(head2.value < head1.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

快速找到链表的中点

• 快慢指针的方式，慢指针每次移动一个长度，快指针每次移动二个长度。当快指针到达链表的末尾，慢指针到达链表的中点。
• 问题：奇数偶数判定不一样的，需要细节定制

回文结构

• 将链表的后半部分放到栈里边，存储的是后半部分数据的逆序
• 面试考虑内存空间，进一步优化。找到链表的中点之后，将中点的下一个指针指向nullptr，然后将中点之后的节点的下一个指针进行逆向排序，如下图所示，主要目的是为了节省空间。然后两个指针一个从头 一个从尾进行依次读取数据进行比较，如果一致满足回文结构。但是最后需要将链表进行还原，恢复成 1 -> 2 -> 3 -> 2 ->1 的结构
• 1 -> 2 -> 3 <- 2 <- 1
• 第二种方式 初始化的时候，将right初始化为head->next；将cur初始化为head节点；主要目的是为了将慢指针移动到中心节点的右边，然后将中心节点开始的位置到null这段区间的节点都拷贝到栈空间

package class04;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 spacepublic 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 spacepublic 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 spacepublic 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 noden1 = n1.next; // n1 -> midn2 = n2.next.next; // n2 -> end}n2 = n1.next; // n2 -> right part first noden1.next = null; // mid.next -> nullNode n3 = null;while (n2 != null) { // right part convertn3 = n2.next; // n3 -> save next noden2.next = n1; // next of right node convertn1 = n2; // n1 moven2 = n3; // n2 move}n3 = n1; // n3 -> save last noden2 = head;// n2 -> left first nodeboolean res = true;while (n1 != null && n2 != null) { // check palindromeif (n1.value != n2.value) {res = false;break;}n1 = n1.next; // left to midn2 = n2.next; // right to mid}n1 = n3.next;n3.next = null;while (n1 != null) { // recover listn2 = n1.next;n1.next = n3;n3 = n1;n1 = n2;}return res;}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 head = null;printLinkedList(head);System.out.print(isPalindrome1(head) + " | ");System.out.print(isPalindrome2(head) + " | ");System.out.println(isPalindrome3(head) + " | ");printLinkedList(head);System.out.println("=========================");head = new Node(1);printLinkedList(head);System.out.print(isPalindrome1(head) + " | ");System.out.print(isPalindrome2(head) + " | ");System.out.println(isPalindrome3(head) + " | ");printLinkedList(head);System.out.println("=========================");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);System.out.println("=========================");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);System.out.println("=========================");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);System.out.println("=========================");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);System.out.println("=========================");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);System.out.println("=========================");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);System.out.println("=========================");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);System.out.println("=========================");}}

将单链表按照给定的数值，划分成左边小、中间相等、右边大的形式 (荷兰国旗)

思路 方法一

• 1，申请一个链表长度大小一致的数组，然后遍历链表的过程中，将每一个点放到数组容器中，在数组里面进行荷兰国旗问题，再把数组容器中的元素都串联起来，形成一个链表返回

• 第一种方法使用的排序 方式思路以及代码如下：
• 设置 small 从左边开始移动，只要small对应位置的元素的数值小于pivot(用户输入的数值)，small就进行移动，同时移动index，使small和index同时向后移动；注意++small index-- ；和初始赋值有关系
• big从右边开始移动；只有index对应位置上的元素的数值 大于 用户输入的数值，就需要交换 index上的数值  和  big指向的数值；将大的数移动到右边，--big，index不要动，接下来检查index的数值
• 如果index 指向的数值  和  用户输入的数值相等，只需要移动index

    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);}}}

思路 方法二

• 2，保证稳定性：根据题目要求设置三个区，小于区、等于区和大于区，每个区间包含两个变量，开始节点和结尾节点，一共设置6个变量，6个变量都指向null；当遇到第一个元素3的时候，它是小于指定元素5的，将小于区域的开始和结束指针都指向3；第二个元素是5，属于等于区，让等于区的start和end都指向第二个元素5；第3个元素也是5，让等于区的end只想第三个元素。以此类推，得到上面的结果。最后将小于区域的end指针链接等于区域的start指针，等于区域的end指针链接大于区域的start指针。
• 需要注意的是，有可能上面6个变量有不存在的情形，需要仔细判别。
• 链表天然具有稳定性  保持元素的前后位置的一致性；相较于方法一，避免了数据的交换

package com.example.algorithm.demo.class04;import org.w3c.dom.Node;public class Code05_SmallerEqualBigger {public static class Node {public int value;public Node next;public Node(int data){this.value = data;}}public static void swap(Node[] nodeArr,int a,int b){Node tmp = nodeArr[a];nodeArr[a] = nodeArr[b];nodeArr[b] = tmp;}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 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 Node listPartition2(Node head,int pivot){Node sH = null;Node sT = null;Node eH = null;Node eT = null;Node bH = null;Node bT = null;Node next = null;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;if (sT != null){sT.next = eH;eT = eT == null ? sT : eT;}if (eT != null){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

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

        //创建节点 填入元素Node cur = head;Node next = null;while (cur != null){next = cur.next;  //指向新创建的节点的下一个cur.next = new Node(cur.value);  //新建节点cur.next.next = next;  //衔接新创建的节点和先前节点，相当于在指定位置插入节点cur = next;  //指向旧的节点}//分离节点Node res = head.next;cur = head;while (cur != null){next = cur.next.next;curCopy = cur.next;cur.next = next;curCopy.next = next != null ? next.next : null;cur = next;}

Codepackage class04;import java.util.HashMap;public class Code06_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 nodewhile (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 randwhile (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;// splitwhile (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);System.out.println("=========================");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 -> 6head.next.rand = head.next.next.next.next.next; // 2 -> 6head.next.next.rand = head.next.next.next.next; // 3 -> 5head.next.next.next.rand = head.next.next; // 4 -> 3head.next.next.next.next.rand = null; // 5 -> nullhead.next.next.next.next.next.rand = head.next.next.next; // 6 -> 4printRandLinkedList(head);res1 = copyListWithRand1(head);printRandLinkedList(res1);res2 = copyListWithRand2(head);printRandLinkedList(res2);printRandLinkedList(head);System.out.println("=========================");}}

• 两个都有环的 链表 ，可以出现三种情况
• 第一种情形，使用其中一个入环节点绕着环走一圈，如果可以找到另外的入环节点就是第三种情形；否则就是第一种情形
• 第二种情形，将第一个入环节点作为判断终止的条件

package class04;public class Code07_FindFirstIntersectNode {public static class Node {public int value;public Node next;public Node(int data) {this.value = data;}}public static Node getIntersectNode(Node head1, Node head2) {if (head1 == null || head2 == null) {return null;}Node loop1 = getLoopNode(head1);Node loop2 = getLoopNode(head2);if (loop1 == null && loop2 == null) {return noLoop(head1, head2);}if (loop1 != null && loop2 != null) {return bothLoop(head1, loop1, head2, loop2);}return null;}public static Node getLoopNode(Node head) {if (head == null || head.next == null || head.next.next == null) {return null;}Node n1 = head.next; // n1 -> slowNode n2 = head.next.next; // n2 -> fastwhile (n1 != n2) {if (n2.next == null || n2.next.next == null) {return null;}n2 = n2.next.next;n1 = n1.next;}n2 = head; // n2 -> walk again from headwhile (n1 != n2) {n1 = n1.next;n2 = n2.next;}return n1;}public static Node noLoop(Node head1, Node head2) {if (head1 == null || head2 == null) {return null;}Node cur1 = head1;Node cur2 = head2;int n = 0;while (cur1.next != null) {n++;cur1 = cur1.next;}while (cur2.next != null) {n--;cur2 = cur2.next;}if (cur1 != cur2) {return null;}cur1 = n > 0 ? head1 : head2;cur2 = cur1 == head1 ? head2 : head1;n = Math.abs(n);while (n != 0) {n--;cur1 = cur1.next;}while (cur1 != cur2) {cur1 = cur1.next;cur2 = cur2.next;}return cur1;}public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {Node cur1 = null;Node cur2 = null;if (loop1 == loop2) {cur1 = head1;cur2 = head2;int n = 0;while (cur1 != loop1) {n++;cur1 = cur1.next;}while (cur2 != loop2) {n--;cur2 = cur2.next;}cur1 = n > 0 ? head1 : head2;cur2 = cur1 == head1 ? head2 : head1;n = Math.abs(n);while (n != 0) {n--;cur1 = cur1.next;}while (cur1 != cur2) {cur1 = cur1.next;cur2 = cur2.next;}return cur1;} else {cur1 = loop1.next;while (cur1 != loop1) {if (cur1 == loop2) {return loop1;}cur1 = cur1.next;}return null;}}public static void main(String[] args) {// 1->2->3->4->5->6->7->nullNode head1 = new Node(1);head1.next = new Node(2);head1.next.next = new Node(3);head1.next.next.next = new Node(4);head1.next.next.next.next = new Node(5);head1.next.next.next.next.next = new Node(6);head1.next.next.next.next.next.next = new Node(7);// 0->9->8->6->7->nullNode head2 = new Node(0);head2.next = new Node(9);head2.next.next = new Node(8);head2.next.next.next = head1.next.next.next.next.next; // 8->6System.out.println(getIntersectNode(head1, head2).value);// 1->2->3->4->5->6->7->4...head1 = new Node(1);head1.next = new Node(2);head1.next.next = new Node(3);head1.next.next.next = new Node(4);head1.next.next.next.next = new Node(5);head1.next.next.next.next.next = new Node(6);head1.next.next.next.next.next.next = new Node(7);head1.next.next.next.next.next.next = head1.next.next.next; // 7->4// 0->9->8->2...head2 = new Node(0);head2.next = new Node(9);head2.next.next = new Node(8);head2.next.next.next = head1.next; // 8->2System.out.println(getIntersectNode(head1, head2).value);// 0->9->8->6->4->5->6..head2 = new Node(0);head2.next = new Node(9);head2.next.next = new Node(8);head2.next.next.next = head1.next.next.next.next.next; // 8->6System.out.println(getIntersectNode(head1, head2).value);}}

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