引言

链表相关的问题几乎都是coding问题，以下是两个简单的链表问题。

一、单链表或双链表如何反转

1.1 单链表的反转操作

给定一个 Node 结构：

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

这是一个单链表结构， 假设初始时，a --> b --> c --> null ：

Node a = new Node(1);
Node b = new Node(2);
Node c = new Node(3);
a.next = b;
b.next = c;
c.next = null;

那么如何实现指针的反转，最终 c --> b --> a --> null ?

代码实现：

public static Node reverseLinkedList(Node head) {Node prev = null;Node next = null;while (head != null) {// 取得下一个Nodenext = head.next;// 改变当前 head 的指针，使其指向 prevhead.next = prev;// 后两步是变换角色，把当前的 head 变成 prevprev = head;// 把 next 下一次循环的 headhead = next;}// 因为最后 head 为 null 才会跳出循环，因此，需要返回 prev 元素return prev;
}

测试：

public static void main(String[] args) {Node a = new Node(1);Node b = new Node(2);Node c = new Node(3);a.next = b;b.next = c;c.next = null;Node head = a;printNodes(head);Node newHead = reverseLinkedList(head);System.out.println("\n反转后：");printNodes(newHead);
}public static void printNodes(Node head) {while (head != null) {System.out.print(head.value);head = head.next;}
}

1.2 双向链表反转

public class LinkedListTest2 {public static Node[] nodes = new Node[3];static {Node a = new Node(1);Node b = new Node(2);Node c = new Node(3);a.prev = null;a.next = b;b.prev = a;b.next = c;c.prev = b;c.next = null;nodes[0] = a;nodes[1] = b;nodes[2] = c;}public static void main(String[] args) {Node head = nodes[0];printNodes(head);Node newHead = reverseLinkedList(head);System.out.println("\n反转后：");printNodes(newHead);}public static void printNodes(Node head) {while (head != null) {System.out.println(head.prev + " <-- " + head.value + " --> " + head.next);head = head.next;}}public static class Node {public Node prev;public int value;public Node next;public Node(int value) {this.value = value;}public String toString() {return Integer.valueOf(value).toString();}}/*** prev <-- head --> next*/public static Node reverseLinkedList(Node head) {Node prev = null;Node next = null;while (head != null) {next = head.next;prev = head.prev;head.prev = next;head.next = prev;prev = head;head = next;}return prev;}
}

二、把给定值的节点都删除

LeetCode 地址：剑指 Offer 18. 删除链表的节点

    /*** 2 -> 2-> 3-> 2-> 3-> 4* 若：删除 num = 3，得到：* 2 -> 2-> 2-> 4，return head = 2* 若：删除 num = 2 得到：* 3-> 3-> 4 ，return head= 3* @param head* @param val* @return*/public static ListNode removeValue(ListNode head, int val) {// head 来到第一个不需要删除的位置while(head != null) {if (head.val != val) {break;}head = head.next;}// 1、head == null 表示所有 node 都是 num，return null// 2、head != nullListNode prev = head;ListNode curr = head;while (curr != null) {if (curr.val == val)prev.next = curr.next;elseprev = curr;curr = curr.next;}return head;}