基礎演算法面試題---連結串列
阿新 • • 發佈:2021-02-12
前言
一般連結串列的基礎題演算法都很簡單,但卻是常見的面試題,因為連結串列能夠考察面試者的編碼能力,往往很容易想到解題方式,卻寫不出來。
下面總結了幾道常見的初級題,可以反覆練習,提高自己的編碼能力。
先準備兩個物件,一個單鏈表,一個雙鏈表
單鏈表
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 +
'}';
}
}
雙鏈表
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、單鏈表反轉
單鏈表反轉,經典且基礎的連結串列題。
舉例:
原連結串列:1-2-3-4-5
反轉後: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("原連結串列:" + head);
ListNode listNode = reverseList(head);
System.out.println("反轉後:" + 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、雙鏈表反轉
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("原連結串列:" + n1);
DoubleListNode listNode = reverseDoubleList(n1);
System.out.println("反轉後:" + 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、刪除連結串列中的某類節點
舉例:
原連結串列:1-2-3-4-5
刪除value=2的節點
結果:1-3-4-5
原連結串列:1-2-3-3-3-4-5
刪除value=3的節點
結果:1-2-4-5
原連結串列:1-1-2-3-4-5
刪除value=1的節點
結果: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("原連結串列:" + head);
ListNode node = deleteNode(head, 3);
System.out.println("刪除後:" + node);
}
private static ListNode deleteNode(ListNode head, int value) {
//處理head本身就是要刪除的節點
while (head != null) {
if (head.val != value) {
break;
}
head = head.next;
}
//始終記錄前一個節點,和當前節點的指標,如果當前節點就是要刪除的節點時,則讓前一個節點指向當前節點的下一個節點,即完成了刪除
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、刪除重複元素1
給定一個排序連結串列,刪除所有重複的元素,使得每個元素只出現一次。
舉例:
原連結串列: 1->1->2->3->3
刪除後: 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("原連結串列:" + head);
System.out.println("刪除後:" + c.deleteDuplicates(head));
}
/**
* 通過不斷移動cur,判斷當前cur的值與cur.next的值是否相等,如果相等,則只改變cur.next,並讓其指向下一個節點,就等於跳過了cur.next的節點
* 如果不相等,則移動cur節點位置到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、刪除重複元素2
給定一個排序連結串列,刪除所有含有重複數字的節點,只保留原始連結串列中沒有重複出現的數字。
舉例:
原連結串列: 1->2->3->3->4->4->5
刪除後: 1->2->5
這是在前一道題目上的延伸。
解法一:
結合前兩題的方式,可以拆分處理,先用第4題的方式刪除重複的數字並記錄下來,再第3題的方式刪除指定數字。
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;
}
}
當然第一種解法只能當做是編碼的練習,出題者肯定不是希望你用這種方式處理。
解法二:
啞節點+雙指標
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));
}
/**
* 啞節點+雙指標
* <p>
* 構建一個啞節點,讓其next指向頭位置。
* 再利用雙指標,n1,n2,初始都指向head位置,如果n1.next.val==n2.next.val,則讓n2向前移動一位,否則n1,n2一起向前移動一位
*
* @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 {
//n2一直移動,直到不等於n1為止
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、合併有序連結串列
將兩個升序連結串列合併為一個新的 升序 連結串列並返回。新連結串列是通過拼接給定的兩個連結串列的所有節點組成的。
原連結串列:l1 = [1,2,4],l2 = [1,3,4]
合併後:[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("原連結串列1:" + l1);
System.out.println("原連結串列2:" + l2);
System.out.println("合併後:" + 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;
//比較兩個連結串列當前的值,值小的連結串列就把引用賦給mergeNode,並向後移動一位重新賦值給自己,同時pre指向值小的那個節點
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、雙向連結串列實現棧、佇列
佇列:先進先出
棧:先進後出
分別用雙向連結串列實現,從頭進,從頭出,從尾進,從尾出即可模擬出佇列和棧的資料結構。
public class NodeUtils<T> {
DoubleListNode<T> head;
DoubleListNode<T> tail;
/**
* 從頭進
* @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;
}
}
/**
* 從尾進
* @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;
}
/**
* 從頭出
* @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;
}
/**
* 從尾出
* @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;
}
}
佇列
/**
* 佇列(先進先出):從連結串列頭進,從連結串列尾出
* @param <T>
*/
public class ListNodeQueue<T> {
NodeUtils<T> nodeUtils = new NodeUtils();
public void push(T data) {
nodeUtils.addHead(data);
}
public T pop() {
return nodeUtils.popTail();
}
}
棧
/**
* 佇列(先進先出):從連結串列頭進,從連結串列尾出
*
* @param <T>
*/
public class ListNodeQueue<T> {
NodeUtils<T> nodeUtils = new NodeUtils();
public void push(T data) {
nodeUtils.addHead(data);
}
public T pop() {
return nodeUtils.popTail();
}
}
驗證
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("棧:壓進去1,2,3");
System.out.println("彈出" + stack.pop() + "," + stack.pop() + "," + stack.pop());
ListNodeQueue queue = new ListNodeQueue();
queue.push(1);
queue.push(2);
queue.push(3);
System.out.println("佇列:壓進去1,2,3");
System.out.println("彈出" + queue.pop() + "," + queue.pop() + "," + queue.pop());
}
}