算法-链表操作

题目

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

思路：快慢指针

// 奇数 arr[len/2]
// 偶数 arr[len/2-1]
// 中间偏上
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 arr[len/2]
// 偶数 arr[len/2]
// 中间偏下
public static Node midOrDownMidNode(Node head) {if (head == null || head.next == null) {return head;}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 arr[len/2-1]
// 偶数 arr[len/2-2]
// 中间偏上前一个
public static Node midOrUpMidPreNode(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 arr[len/2-1]
// 偶数 arr[len/2-1]
// 中间偏上后一个
public static Node midOrDownMidPreNode(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;}

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。

1）哈希表方法特别简单（笔试用）

2）改原链表的方法就需要注意边界了（面试用）

思路：快慢指针，将中间元素next指针指向null ，同时更不后半部分next指向前一节点。前后双指正向中间遍历，直到为任意指针为null，时间复杂度n ，空间复杂度1

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 noden1 = n1.next; // n1 -> midn2 = n2.next.next; // n2 -> end}// n1 中点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;}

思路：空间换时间将来链表后半部放入堆栈,空间复杂度O(n/2)

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;}

题目：将单向链表按某值划分成左边小、中间相等、右边大的形式

思路：把链表放入数组里，在数组上做partition

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;}

思路：分成小、中、大三部分，再把各个部分之间串起来

public static Node listPartition2(Node head, int pivot) {if (head == null)return null;Node lesshead = new Node(0);Node lesstail = lesshead;Node equalhead = new Node(0);Node equaltail = equalhead;Node morehead = new Node(0);Node moretail = morehead;Node cur = head;while (cur != null){if(cur.value == pivot){equaltail.next = cur;equaltail = equaltail.next;}else if (cur.value<pivot){lesstail.next = cur;lesstail = lesstail.next;}else{moretail = cur;moretail = moretail.next;}cur = cur.next;}// 关键步骤在这里equaltail.next = morehead.next;lesstail.next = equalhead.next;return lesshead.next;}