一、链表问题

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

二、快慢指针

逻辑：
慢指针一次走1步
快指针一次走2步
当快指针走完的时候，慢指针应该来到中点的位置

1、输入链表头节点，奇数长度返回中点，偶数长度返回上中点

2、输入链表头节点，奇数长度返回中点，偶数长度返回下中点

3、输入链表头节点，奇数长度返回中点前一个，偶数长度返回上中点前一个

4、输入链表头节点，奇数长度返回中点前一个，偶数长度返回下中点前一个

package class06;import java.util.ArrayList;/*** 快慢指针*/
public class Code01_LinkedListMid {public static class Node {public int value;public Node next;public Node(int v) {value = v;}}/*** 奇数情况返回中点，偶数情况返回上中点* @param head：头节点* @return*/public static Node midOrUpMidNode(Node head) {if (head == null || head.next == null || head.next.next == null) {return head;}// 链表有3个点或以上Node slow = head.next;Node fast = head.next.next;while (fast.next != null && fast.next.next != null) {slow = slow.next;fast = fast.next.next;}return slow;}/*** 奇数情况返回中点，偶数情况返回下中点* @param head：头节点* @return*/public static Node midOrDownMidNode(Node head) {if (head == null || head.next == null) {return head;}// 初始设置不一样Node slow = head.next;Node fast = head.next;while (fast.next != null && fast.next.next != null) {slow = slow.next;fast = fast.next.next;}return slow;}/*** 奇数情况返回中点前一个，偶数情况返回上中点前一个* @param head：头节点* @return*/public static Node midPreOrUpMidPreNode(Node head) {if (head == null || head.next == null || head.next.next == null) {return null;}Node slow = head;Node fast = head.next.next;while (fast.next != null && fast.next.next != null) {slow = slow.next;fast = fast.next.next;}return slow;}/*** 奇数情况返回中点前一个，偶数情况返回下中点前一个* @param head：头节点* @return*/public static Node midPreOrDownMidPreNode(Node head) {if (head == null || head.next == null) {return null;}if (head.next.next == null) {return head;}Node slow = head;Node fast = head.next;while (fast.next != null && fast.next.next != null) {slow = slow.next;fast = fast.next.next;}return slow;}public static Node right1(Node head) {if (head == null) {return head;}Node cur = head;ArrayList<Node> arr = new ArrayList<>();while (cur != null) {arr.add(cur);cur = cur.next;}return arr.get((arr.size() - 1) / 2);}public static Node right2(Node head) {if (head == null) {return null;}Node cur = head;ArrayList<Node> arr = new ArrayList<>();while (cur != null) {arr.add(cur);cur = cur.next;}return arr.get(arr.size() / 2);}public static Node right3(Node head) {if (head == null || head.next == null || head.next.next == null) {return null;}Node cur = head;ArrayList<Node> arr = new ArrayList<>();while (cur != null) {arr.add(cur);cur = cur.next;}return arr.get((arr.size() - 3) / 2);}public static Node right4(Node head) {if (head == null || head.next == null) {return null;}Node cur = head;ArrayList<Node> arr = new ArrayList<>();while (cur != null) {arr.add(cur);cur = cur.next;}return arr.get((arr.size() - 2) / 2);}public static void main(String[] args) {Node head = new Node(0);Node tail = head;for (int i = 1; i < 10; i++) {Node node = new Node(i);tail.next = node;tail = tail.next;}System.out.println(midOrUpMidNode(head).value);System.out.println(right1(head).value);System.out.println(midOrDownMidNode(head).value);System.out.println(right2(head).value);System.out.println(midPreOrUpMidPreNode(head).value);System.out.println(right3(head).value);System.out.println(midPreOrDownMidPreNode(head).value);System.out.println(right4(head).value);}
}

三、面试题1

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
回文结构：正着念和返着念都一样，例如：12aa21、12321

1、栈的方法特别简单（笔试用）
1 -> 2 -> 3 -> 2 -> 1
第一遍遍历：一个一个放到栈里去
第二遍遍历：从栈里弹出一个（是倒叙），和第一个比较，以此类推

2、改原链表的方法就需要注意边界了（面试用）

1 -> 2 -> 3 <- 2 <- 1
               |
               |-> null

package class06;import java.util.Stack;import class06.Code01_LinkedListMid.Node;public class Code02_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>(); //准备一个栈Node cur = head; //引用从头节点开始while (cur != null) {stack.push(cur); //所有节点加到栈里去cur = cur.next;}while (head != null) {//从头开始遍历，和栈中弹出的值比较if (head.value != stack.pop().value) {return false;}head = head.next;}return true;}// need n/2 extra space（省一点空间的方法）public static boolean isPalindrome2(Node head) {if (head == null || head.next == null) {return true;}//使用快慢指针，奇数定位到唯一终点，偶数定位到上中点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); //把右半部分加到栈里去right = right.next;}while (!stack.isEmpty()) {if (head.value != stack.pop().value) {return false;}head = head.next;}return true;}//不用容器的方法//把有右半部分逆序，左指针和右指针比较//need O(1) extra spacepublic static boolean isPalindrome3(Node head) {if (head == null || head.next == null) {return true;}Node n1 = head; //慢指针Node n2 = head; //快指针//找中点while (n2.next != null && n2.next.next != null) { // find mid noden1 = n1.next; // n1 -> midn2 = n2.next.next; // n2 -> end}//右半部分逆序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 node//比较左右指针boolean 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}//右半部分还原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 main(String[] args) {int[] arr = {1, 2, 3, 3, 2, 1};Node head = new Node(1);Node tail = head;for (int i = 1; i < arr.length; i++) {Node node = new Node(arr[i]);tail.next = node;tail = tail.next;}System.out.println(isPalindrome1(head));System.out.println(isPalindrome2(head));System.out.println(isPalindrome3(head));}
}