基础算法面试题---链表
前言
一般链表的基础题算法都很简单，但却是常见的面试题，因为链表能够考察面试者的编码能力，往往很容易想到解题方式，却写不出来。
下面总结了几道常见的初级题，可以反复练习，提高自己的编码能力。
先准备两个对象，一个单链表，一个双链表
单链表
public class ListNode {
int val;
ListNode next;
ListNode() {
}
ListNode(int val) {
this.val = val;
}
ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
@Override
public String toString() {
return "ListNode{" +
"val=" + val +
", next=" + next +
'}';
}
}
双链表
public class DoubleListNode<T> {
public T data;
public DoubleListNode<T> last;
public DoubleListNode<T> next;
public DoubleListNode(T data) {
this.data = data;
}
@Override
public String toString() {
return "DoubleListNode{" +
"data=" + data +
", next=" + next +
'}';
}
}
1、单链表反转
单链表反转，经典且基础的链表题。
举例：
原链表：1-2-3-4-5
反转后：5-4-3-2-1
public class Code_01 {
public static void main(String[] args) {
ListNode n = new ListNode();
ListNode head = n;
for (int i = 1; i <= 5; i++) {
n.next = new ListNode(i);
n = n.next;
}
System.out.println("原链表：" + head);
ListNode listNode = reverseList(head);
System.out.println("反转后：" + listNode);
}
private static ListNode reverseList(ListNode head) {
ListNode pre = null;
ListNode next;
while (head != null) {
next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
}
2、双链表反转
public class Code_02 {
public static void main(String[] args) {
DoubleListNode n1 = new DoubleListNode(1);
DoubleListNode n2 = new DoubleListNode(2);
n1.next = n2;
n2.last = n1;
DoubleListNode n3 = new DoubleListNode(3);
n2.next = n3;
n3.last = n2;
System.out.println("原链表：" + n1);
DoubleListNode listNode = reverseDoubleList(n1);
System.out.println("反转后：" + listNode);
}
private static DoubleListNode reverseDoubleList(DoubleListNode head) {
DoubleListNode pre = null;
DoubleListNode next;
while (head != null) {
next = head.next;
head.next = pre;
head.last = next;
pre = head;
head = next;
}
return pre;
}
}
3、删除链表中的某类节点
举例：
原链表：1-2-3-4-5
删除value=2的节点
结果：1-3-4-5
原链表：1-2-3-3-3-4-5
删除value=3的节点
结果：1-2-4-5
原链表：1-1-2-3-4-5
删除value=1的节点
结果：2-3-4-5
public class Code_03 {
public static void main(String[] args) {
ListNode n = new ListNode();
ListNode head = n;
for (int i = 1; i <= 5; i++) {
n.next = new ListNode(i);
n = n.next;
}
System.out.println("原链表：" + head);
ListNode node = deleteNode(head, 3);
System.out.println("删除后：" + node);
}
private static ListNode deleteNode(ListNode head, int value) {
//处理head本身就是要删除的节点
while (head != null) {
if (head.val != value) {
break;
}
head = head.next;
}
//始终记录前一个节点，和当前节点的指针，如果当前节点就是要删除的节点时，则让前一个节点指向当前节点的下一个节点，即完成了删除
ListNode cur = head;
ListNode pre = null;
while (cur != null) {
if (cur.val == value) {
pre.next = cur.next;
} else {
pre = cur;
}
cur = cur.next;
}
return head;
}
}
4、删除重复元素1
给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。
举例：
原链表: 1->1->2->3->3
删除后: 1->2->3
public class Code_05 {
public static void main(String[] args) {
Code_05 c = new Code_05();
ListNode n = new ListNode();
ListNode head = n;
for (int i = 1; i <= 5; i++) {
n.next = new ListNode(i);
n = n.next;
if (i % 2 == 0) {
n.next = new ListNode(i);
n = n.next;
}
}
System.out.println("原链表：" + head);
System.out.println("删除后：" + c.deleteDuplicates(head));
}
/**
* 通过不断移动cur，判断当前cur的值与cur.next的值是否相等，如果相等，则只改变cur.next，并让其指向下一个节点，就等于跳过了cur.next的节点
* 如果不相等，则移动cur节点位置到cur.next上。
* @param head
* @return
*/
public ListNode deleteDuplicates(ListNode head) {
ListNode cur = head;
while (cur != null && cur.next != null) {
if (cur.val != cur.next.val) {
cur = cur.next;
} else {
cur.next = cur.next.next;
}
}
return head;
}
}
5、删除重复元素2
给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中没有重复出现的数字。
举例：
原链表: 1->2->3->3->4->4->5
删除后: 1->2->5
这是在前一道题目上的延伸。
解法一：
结合前两题的方式，可以拆分处理，先用第4题的方式删除重复的数字并记录下来，再第3题的方式删除指定数字。
public class Code_06 {
public static void main(String[] args) {
Code_06 c = new Code_06();
ListNode n = new ListNode();
ListNode head = n;
for (int i = 1; i <= 5; i++) {
n.next = new ListNode(i);
n = n.next;
if (i % 2 == 0) {
n.next = new ListNode(i);
n = n.next;
}
}
System.out.println(head);
System.out.println(c.deleteDuplicates(head));
}
public ListNode deleteDuplicates(ListNode head) {
ListNode cur = head;
Set<Integer> dupSet = new HashSet<>();
while (cur != null && cur.next != null) {
if (cur.val != cur.next.val) {
cur = cur.next;
} else {
dupSet.add(cur.val);
cur.next = cur.next.next;
}
}
for (Integer val : dupSet) {
head = deleteNode(head, val);
}
return head;
}
private ListNode deleteNode(ListNode head, int val) {
while (head != null) {
if (head.val != val) {
break;
}
head = head.next;
}
ListNode pre = null;
ListNode cur = head;
while (cur != null) {
if (cur.val == val) {
pre.next = cur.next;
} else {
pre = cur;
}
cur = cur.next;
}
return head;
}
}
当然第一种解法只能当做是编码的练习，出题者肯定不是希望你用这种方式处理。
解法二：
哑节点+双指针
public class Code_06_01 {
public static void main(String[] args) {
Code_06_01 c = new Code_06_01();
ListNode n = new ListNode();
ListNode head = n;
for (int i = 1; i <= 5; i++) {
n.next = new ListNode(i);
n = n.next;
if (i % 2 == 0) {
n.next = new ListNode(i);
n = n.next;
}
}
System.out.println(head);
System.out.println(c.deleteDuplicates(head));
}
/**
* 哑节点+双指针
*
* 构建一个哑节点，让其next指向头位置。
* 再利用双指针，n1,n2,初始都指向head位置，如果n1.next.val==n2.next.val，则让n2向前移动一位，否则n1，n2一起向前移动一位
*
* @param head
* @return
*/
public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = new ListNode();
dummy.next = head;
ListNode n1 = dummy;
ListNode n2 = head;
while (n2 != null && n2.next != null) {
if (n1.next.val != n2.next.val) {
n1 = n1.next;
} else {
//n2一直移动，直到不等于n1为止
while (n2 != null && n2.next != null && n1.next.val == n2.next.val) {
n2 = n2.next;
}
n1.next = n2.next;
}
n2 = n2.next;
}
return dummy.next;
}
}
6、合并有序链表
将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
原链表：l1 = [1,2,4]，l2 = [1,3,4]
合并后：[1,1,2,3,4,4]
public class Code_04 {
public static void main(String[] args) {
ListNode n = new ListNode(1);
ListNode l1 = n;
for (int i = 3; i <= 5; i = i + 2) {
n.next = new ListNode(i);
n = n.next;
}
ListNode n2 = new ListNode(2);
ListNode l2 = n2;
for (int i = 4; i <= 6; i = i + 2) {
n2.next = new ListNode(i);
n2 = n2.next;
}
Code_04 c = new Code_04();
System.out.println("原链表1：" + l1);
System.out.println("原链表2：" + l2);
System.out.println("合并后：" + c.mergeTwoLists(l1, l2));
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
ListNode mergeNode = new ListNode();
ListNode pre = mergeNode;
//比较两个链表当前的值，值小的链表就把引用赋给mergeNode，并向后移动一位重新赋值给自己，同时pre指向值小的那个节点
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
pre.next = l1;
l1 = l1.next;
} else {
pre.next = l2;
l2 = l2.next;
}
pre = pre.next;
}
pre.next = l1 == null ? l2 : l1;
return mergeNode.next;
}
}
7、双向链表实现栈、队列
队列：先进先出
栈：先进后出
分别用双向链表实现，从头进，从头出，从尾进，从尾出即可模拟出队列和栈的数据结构。
public class NodeUtils<T> {
DoubleListNode<T> head;
DoubleListNode<T> tail;
/**
* 从头进
* @param data
*/
public void addHead(T data) {
DoubleListNode<T> node = new DoubleListNode(data);
if (head == null) {
head = node;
tail = node;
} else {
node.next = head;
head.last = node;
head = node;
}
}
/**
* 从尾进
* @param data
*/
public void addTail(T data) {
DoubleListNode<T> node = new DoubleListNode(data);
if (head == null) {
head = node;
} else {
tail.next = node;
node.last = tail;
}
tail = node;
}
/**
* 从头出
* @return
*/
public T popHead() {
if (head == null) {
return null;
}
DoubleListNode<T> h = head;
if (head == tail) {
head = null;
tail = null;
} else {
head = head.next;
head.last = null;
}
return h.data;
}
/**
* 从尾出
* @return
*/
public T popTail() {
if (tail == null) {
return null;
}
DoubleListNode<T> t = tail;
if (head == tail) {
head = null;
tail = null;
} else {
tail = tail.last;
tail.next = null;
}
return t.data;
}
}
队列
/**
* 队列（先进先出）：从链表头进，从链表尾出
* @param
*/
public class ListNodeQueue<T> {
NodeUtils<T> nodeUtils = new NodeUtils();
public void push(T data) {
nodeUtils.addHead(data);
}
public T pop() {
return nodeUtils.popTail();
}
}
栈
/**
* 队列（先进先出）：从链表头进，从链表尾出
*
* @param
*/
public class ListNodeQueue<T> {
NodeUtils<T> nodeUtils = new NodeUtils();
public void push(T data) {
nodeUtils.addHead(data);
}
public T pop() {
return nodeUtils.popTail();
}
}
验证
public class Code_07 {
public static void main(String[] args) {
ListNodeStack stack = new ListNodeStack();
stack.push(1);
stack.push(2);
stack.push(3);
System.out.println("栈：压进去1,2,3");
System.out.println("弹出" + stack.pop() + "," + stack.pop() + "," + stack.pop());
ListNodeQueue queue = new ListNodeQueue();
queue.push(1);
queue.push(2);
queue.push(3);
System.out.println("队列：压进去1,2,3");
System.out.println("弹出" + queue.pop() + "," + queue.pop() + "," + queue.pop());
}
}
