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java面试编程面试题
Java Programming Interview Questions are always the deciding factor in any Java interview. Whether you are a beginner in Java or an expert programmer, you will be tested for your coding skills in the interview. So, it’s a good idea to brush up your coding skills before you face the interview.
Java编程面试问题始终是任何Java面试的决定因素。 无论您是Java的初学者还是专家程序员，都将在面试中测试您的编码技能。 因此，在面试之前提高自己的编码技能是一个好主意。
I am providing 50+ Java programming questions to test your coding skills. It’s good for beginners as well as experienced Java developers. The best part is that some of the questions are from the latest releases (Java 14). If you can write Java programs using the latest features, it means you keep yourself up to date, which will work in your favor.
我提供了50多个Java编程问题来测试您的编码技能。 对于初学者和有经验的Java开发人员来说都非常好。 最好的部分是一些问题来自最新版本（Java 14）。 如果您可以使用最新功能编写Java程序，则意味着您可以保持最新状态，这将对您有利。
It might be surprising, but there is no reverse() utility method in the String class. But, it’s a very simple task. We can create a character array from the string and then iterate it from the end to start. We can append the characters to a string builder and finally return the reversed string.
可能令人惊讶，但是String类中没有反向（）实用程序方法。 但是，这是一个非常简单的任务。 我们可以从字符串创建一个字符数组，然后从头开始迭代它。 我们可以将字符附加到字符串生成器，最后返回反转的字符串。
public class StringPrograms { public static void main(String[] args) { String str = "123"; System.out.println(reverse(str)); } public static String reverse(String in) { if (in == null) throw new IllegalArgumentException("Null is not valid input"); StringBuilder out = new StringBuilder(); char[] chars = in.toCharArray(); for (int i = chars.length - 1; i > 0; i--) out.append(chars[i]); return out.toString(); }} Bonus Points: Adding null check in the method and using StringBuilder for appending the characters.
优点 ：在方法中添加null检查，并使用StringBuilder附加字符。
Easy to Miss: The indexing in Java starts from 0. So, we have to start at “chars.length – 1” in the for loop.
容易错过 ：Java中的索引从0开始。因此，我们必须在for循环中从“ chars.length – 1”开始。
It’s a slightly tricky question. It’s a three steps process. It’s better visualized in code.
这是一个有点棘手的问题。 这是一个三步过程。 用代码更好地可视化。
int a = 10;int b = 20;b = b + a; // now b is sum of both the numbersa = b - a; // b - a = (b + a) - a = b (a is swapped)b = b - a; // (b + a) - b = a (b is swapped)
We can’t return multiple variables in Java. Since Java is Pass-by-Value and these are primitive data types, their values won’t change. For example, below swap function will not change the input integer values.
我们无法在Java中返回多个变量。 由于Java是“按值传递”并且它们是原始数据类型，因此它们的值不会更改。 例如，下面的swap函数将不会更改输入整数值。
public static void swapNumbers(int a, int b) { b = b + a; a = b - a; b = b - a;}public static void main(String[] args) { int a = 10; int b = 20; swapNumbers(a, b); System.out.printf("a is %d, b is %d", a, b); // a is 10, b is 20} We can use regular expression to check if the string contains vowels or not.
我们可以使用正则表达式来检查字符串是否包含元音。
public class StringContainsVowels { public static void main(String[] args) { System.out.println(stringContainsVowels("Hello")); // true System.out.println(stringContainsVowels("TV")); // false } public static boolean stringContainsVowels(String input) { return input.toLowerCase().matches(".*[aeiou].*"); }} We can write a simple program to divide the given number “n” from 2 to n/2 and check the remainder. If the remainder is 0, then it’s not a prime number.
我们可以编写一个简单的程序将给定的数字“ n”从2除以n / 2，然后检查余数。 如果余数为0，则它不是质数。
public class PrimeNumberCheck { public static void main(String[] args) { System.out.println(isPrime(19)); // true System.out.println(isPrime(49)); // false } public static boolean isPrime(int n) { if (n == 0 || n == 1) { return false; } if (n == 2) { return true; } for (int i = 2; i <= n / 2; i++) { if (n % i == 0) { return false; } } return true; }} But, this is not very memory and time-efficient. For a given number N, if there is a prime number M between 2 to √N (square root of N) that evenly divides it, then N is not a prime number.
但是，这不是很节省内存和时间。 对于给定的数N，如果在2到√N（N的平方根）之间有一个质数M对其进行均分，则N不是质数。
You can read more about it .
您可以了解更多信息。
We can use a for loop to print fibonacci series.
我们可以使用for循环来打印斐波那契数列。
public static void printFibonacciSeries(int count) { int a = 0; int b = 1; int c = 1; for (int i = 1; i <= count; i++) { System.out.print(a + ", "); a = b; b = c; c = a + b; }} The fibonacci number is generated by adding the previous two numbers – F(N) = F(N-1) + F(N-2). We can use recursion to print fibonacci series.
斐波那契数是通过将前两个数相加而得出的-F（N）= F（N-1）+ F（N-2）。 我们可以使用递归来打印斐波那契数列。
public class FibonacciNumbers { public static int fibonacci(int n) { if (n <= 1) return n; return fibonacci(n - 1) + fibonacci(n - 2); } public static void main(String args[]) { int n = 10; System.out.println(fibonacci(n)); }} We can use for loop and check each element one by one if they are odd or not.
我们可以使用for循环并逐个检查每个元素是否为奇数。
public static boolean onlyOddNumbers(Listlist) { for (int i : list) { if (i % 2 == 0) return false; } return true;}
If the list is huge, we can use parallel stream for faster processing.
如果列表很大，我们可以使用并行流进行更快的处理。
public static boolean onlyOddNumbers(Listlist) { return list .parallelStream() // parallel stream for faster processing .anyMatch(x -> x % 2 != 0); // return as soon as any elements match the condition}
A palindrome string is one whose reverse is also the same string. So we can reverse the input string and check if both strings are equal or not. We can also use the String charAt(int index) method to check for palindrome string.
回文字符串是反向字符串也相同的字符串。 因此，我们可以反转输入字符串，并检查两个字符串是否相等。 我们还可以使用String charAt（int index）方法检查回文字符串。
boolean checkPalindromeString(String input) { boolean result = true; int length = input.length(); for(int i=0; i < length/2; i++) { if(input.charAt(i) != input.charAt(length-i-1)) { result = false; break; } } return result;} We can use Character.isWhitespace() method to remove whitespaces from the string.
我们可以使用Character.isWhitespace（）方法从字符串中删除空格。
String removeWhiteSpaces(String input){ StringBuilder output = new StringBuilder(); char[] charArray = input.toCharArray(); for(char c : charArray) { if (!Character.isWhitespace(c)) output.append(c); } return output.toString();} Java String class contains two methods to remove leading and trailing whitespaces – trim(), and strip(). The strip() method was added to the String class in Java 11. However, the strip() method uses Character.isWhitespace() method to check if the character is a whitespace. This method uses Unicode code points whereas the trim() method identifies any character having codepoint value less than or equal to ‘U+0020’ as a whitespace character.
Java String类包含两种删除开头和结尾空格的方法– trim（）和strip（）。 strip（）方法已添加到Java 11中的String类中。但是，strip（）方法使用Character.isWhitespace（）方法来检查字符是否为空格。 此方法使用Unicode代码点，而trim（）方法将代码点值小于或等于'U + 0020'的任何字符标识为空白字符。
The strip() method is the recommended way to remove whitespaces because it uses the Unicode standard.
推荐使用strip（）方法删除空格，因为它使用Unicode标准。
String s = " abc def\t"; s = s.strip(); System.out.println(s);
Since String is immutable, we have to assign the strip() output to the string.
由于String是不可变的，因此我们必须将strip（）输出分配给字符串。
This question requires a deep understanding of sorting in Java. If you look at the Arrays utility class, there are many overloaded sort() methods to sort primitive as well as to object arrays.
这个问题需要对Java排序有深入的了解。 如果您看一下Arrays实用程序类，有很多重载的sort（）方法可以对基元和对象数组进行排序。
If you are sorting a primitive array in the natural order, then it’s very simple. Just use the Arrays.sort() method.
如果您要按自然顺序对原始数组进行排序，那么这很简单。 只需使用Arrays.sort（）方法即可。
int[] array = {1, 2, 3, -1, -2, 4};Arrays.sort(array);System.out.println(Arrays.toString(array)); But, if you want to sort an array of Objects, then the object must implement Comparable interface. If you want to specify the sorting criteria, then you can pass the Comparator for the sorting logic. You should read more about them at – .
但是，如果要对对象数组进行排序，则该对象必须实现Comparable接口。 如果要指定排序条件，则可以将比较器传递给排序逻辑。 您应该在– 阅读有关它们的更多信息。
Deadlock is a special scenario in the multi-threaded environment where two or more threads are blocked forever. The deadlock situation arises with at least two threads and two or more threads. Let’s write a simple program to create a deadlock.
死锁是多线程环境中的一种特殊情况，在这种情况下，两个或更多线程将永远被阻塞。 至少有两个线程以及两个或更多线程会出现死锁情况。 让我们编写一个简单的程序来创建死锁。
public class ThreadDeadlock { public static void main(String[] args) throws InterruptedException { Object obj1 = new Object(); Object obj2 = new Object(); Object obj3 = new Object(); Thread t1 = new Thread(new SyncThread(obj1, obj2), "t1"); Thread t2 = new Thread(new SyncThread(obj2, obj3), "t2"); Thread t3 = new Thread(new SyncThread(obj3, obj1), "t3"); t1.start(); Thread.sleep(5000); t2.start(); Thread.sleep(5000); t3.start(); }}class SyncThread implements Runnable{ private Object obj1; private Object obj2; public SyncThread(Object o1, Object o2){ this.obj1=o1; this.obj2=o2; } @Override public void run() { String name = Thread.currentThread().getName(); System.out.println(name + " acquiring lock on "+obj1); synchronized (obj1) { System.out.println(name + " acquired lock on "+obj1); work(); System.out.println(name + " acquiring lock on "+obj2); synchronized (obj2) { System.out.println(name + " acquired lock on "+obj2); work(); } System.out.println(name + " released lock on "+obj2); } System.out.println(name + " released lock on "+obj1); System.out.println(name + " finished execution."); } private void work() { try { Thread.sleep(30000); } catch (InterruptedException e) { e.printStackTrace(); } }} All the three threads will be able to acquire a lock on the first object. But, they are using the shared resources and started in such a way that they will keep on waiting indefinitely to acquire the lock on the second object. We can use the java thread dump to detect the deadlocks. Read more at .
这三个线程都将能够获得第一个对象的锁。 但是，他们使用共享资源并以一种方式启动，即他们将无限期地等待获取第二个对象的锁。 我们可以使用java线程转储来检测死锁。 阅读更多内容。
The factorial of an integer is calculated by multiplying all the numbers from 1 to the given number.
整数的阶乘是通过将所有数字从1乘以给定数字来计算的。
F(n) = F(1)*F(2)…F(n-1)*F(n).
F（n）= F（1）* F（2）... F（n-1）* F（n）。
We can use recursion to find the factorial of an integer.
我们可以使用递归来查找整数的阶乘。
public static long factorial(long n) { if (n == 1) return 1; else return (n * factorial(n - 1));} LinkedList descendingIterator() returns an iterator that iterates over the element in the reverse order. We can use this iterator to create a new Linked List with elements in the reverse order.
LinkedListDescendingIterator（）返回一个迭代器，该迭代器以相反的顺序迭代元素。 我们可以使用此迭代器来创建一个新的链表，其中的元素顺序相反。
LinkedListll = new LinkedList<>();ll.add(1);ll.add(2);ll.add(3);System.out.println(ll);LinkedList ll1 = new LinkedList<>();ll.descendingIterator().forEachRemaining(ll1::add);System.out.println(ll1);
If you are looking from the data structures and algorithms perspective, read .
如果您从数据结构和算法的角度看，请阅读 。
The array elements must be sorted for implementing binary search. The binary search algorithm is based on the following conditions.
必须对数组元素进行排序才能实现二进制搜索。 二进制搜索算法基于以下条件。
If the key is less than the middle element, then we now need to search only in the first half of the array.
If the key is greater than the middle element, then we need to only search in the second half of the array. And if the key is equal to the middle element in the array, then the search ends. Finally, if the key is not found in the whole array, then it should return -1. This indicates that the element is not present.如果键小于中间元素，那么我们现在只需要在数组的前半部分搜索。
如果键大于中间元素，那么我们只需要在数组的后半部分搜索即可。 如果键等于数组中的中间元素，则搜索结束。 最后，如果在整个数组中找不到键，则它应该返回-1。 这表明该元素不存在。public static int binarySearch(int arr[], int low, int high, int key) { int mid = (low + high) / 2; while (low <= high) { if (arr[mid] < key) { low = mid + 1; } else if (arr[mid] == key) { return mid; } else { high = mid - 1; } mid = (low + high) / 2; } if (low > high) { return -1; } return -1;} Merge sort is one of the most efficient sorting algorithms. It works on the principle of Divide and Conquers. It is based on the idea of breaking down a list into several sub-lists until each sublist consists of a single element. Then merging those sublists in a manner that results in a sorted list.
合并排序是最有效的排序算法之一。 它遵循分而治之的原则。 它基于以下思想：将一个列表分解为几个子列表，直到每个子列表包含一个元素。 然后以导致排序列表的方式合并这些子列表。
public class MergeSort { public static void main(String[] args) { int[] arr = { 70, 50, 30, 10, 20, 40, 60 }; int[] merged = mergeSort(arr, 0, arr.length - 1); for (int val : merged) { System.out.print(val + " "); } } public static int[] mergeTwoSortedArrays(int[] one, int[] two) { int[] sorted = new int[one.length + two.length]; int i = 0; int j = 0; int k = 0; while (i < one.length && j < two.length) { if (one[i] < two[j]) { sorted[k] = one[i]; k++; i++; } else { sorted[k] = two[j]; k++; j++; } } if (i == one.length) { while (j < two.length) { sorted[k] = two[j]; k++; j++; } } if (j == two.length) { while (i < one.length) { sorted[k] = one[i]; k++; i++; } } return sorted; } public static int[] mergeSort(int[] arr, int lo, int hi) { if (lo == hi) { int[] br = new int[1]; br[0] = arr[lo]; return br; } int mid = (lo + hi) / 2; int[] fh = mergeSort(arr, lo, mid); int[] sh = mergeSort(arr, mid + 1, hi); int[] merged = mergeTwoSortedArrays(fh, sh); return merged; }} Pattern programs are used a lot in interviews to understand the logical thinking abilities of the interviewee. Pyramid patterns are very popular and once we get the logic on the way it’s created, writing code to achieve the same is an easy task.
访谈中经常使用模式程序来了解受访者的逻辑思维能力。 金字塔模式非常流行，一旦我们了解了创建方式的逻辑，编写代码即可实现这一目标很容易。
I have written an extensive post for different kind of pyramid patterns examples, read it .
我为各种不同的金字塔图案示例撰写了广泛的文章，请在阅读。
We will first create a set of elements from both the arrays. Then compare the elements in these sets to find if there is an element that is not present in both the sets?
我们将首先从这两个数组创建一组元素。 然后比较这些集合中的元素以查找两个集合中都不存在的元素？
package com.journaldev.programminginterviews;import java.util.Arrays;import java.util.HashSet;import java.util.Set;public class ArraySameElements { public static void main(String[] args) { Integer[] a1 = {1,2,3,2,1}; Integer[] a2 = {1,2,3}; Integer[] a3 = {1,2,3,4}; System.out.println(sameElements(a1, a2)); System.out.println(sameElements(a1, a3)); } static boolean sameElements(Object[] array1, Object[] array2) { Set It’s a very simple program. We can use for loop to iterate over the array elements and add them to get the final sum.
这是一个非常简单的程序。 我们可以使用for循环遍历数组元素并将其相加以获得最终总和。
int[] array = { 1, 2, 3, 4, 5 };int sum = 0;for (int i : array) sum += i;System.out.println(sum); There are many ways to solve this problem. We can sort the array in natural ascending order and take the second last value. But, sorting is an expensive operation.
有很多方法可以解决此问题。 我们可以按自然升序对数组进行排序，并取倒数第二个值。 但是，分类是一项昂贵的操作。
We can also use two variables to find the second largest value in a single iteration.
我们还可以使用两个变量在一次迭代中找到第二个最大值。
private static int findSecondHighest(int[] array) { int highest = Integer.MIN_VALUE; int secondHighest = Integer.MIN_VALUE; for (int i : array) { if (i > highest) { secondHighest = highest; highest = i; } else if (i > secondHighest) { secondHighest = i; } } return secondHighest;} We can use Random class to generate random index numbers and shuffle the elements.
我们可以使用Random类来生成随机索引号并改组元素。
int[] array = { 1, 2, 3, 4, 5, 6, 7 };Random rand = new Random();for (int i = 0; i < array.length; i++) { int randomIndexToSwap = rand.nextInt(array.length); int temp = array[randomIndexToSwap]; array[randomIndexToSwap] = array[i]; array[i] = temp;}System.out.println(Arrays.toString(array)); We can run the shuffling code inside another for loop to shuffle multiple rounds.
我们可以在另一个for循环中运行改组代码以改组多个回合。
We can use Scanner class to read the file contents line by line. Then use String contains() method to check if the string is present in the file or not.
我们可以使用Scanner类逐行读取文件内容。 然后使用String contains（）方法检查文件中是否存在该字符串。
boolean findStringInFile(String filePath, String str) throws FileNotFoundException { File file = new File(filePath); Scanner scanner = new Scanner(file); // read the file line by line while (scanner.hasNextLine()) { String line = scanner.nextLine(); if (line.contains(str)) { scanner.close(); return true; } } scanner.close(); return false;} The above code assumes that the string we are searching for in the file doesn’t contain newline characters.
上面的代码假定我们在文件中搜索的字符串不包含换行符。
We can use SimpleDateFormat class to get the date string into specific formatting.
我们可以使用SimpleDateFormat类将日期字符串转换为特定格式。
String pattern = "MM-dd-yyyy";SimpleDateFormat simpleDateFormat = new SimpleDateFormat(pattern);String date = simpleDateFormat.format(new Date());System.out.println(date); // 06-23-2020
Recommended Reading: .
推荐读物： 。
We can use the addAll() method to merge multiple lists in Java.
我们可以使用addAll（）方法合并Java中的多个列表。
Listlist1 = new ArrayList<>();list1.add("1");List list2 = new ArrayList<>();list2.add("2");List mergedList = new ArrayList<>(list1);mergedList.addAll(list2);System.out.println(mergedList); // [1, 2]
HashMap is not an ordered collection. So, sorting its entries doesn’t make any sense. But, we can sort the entries based on value and store into LinkedHashMap. LinkedHashMap maintains the order of insertion.
HashMap不是有序集合。 因此，对条目进行排序没有任何意义。 但是，我们可以基于值对条目进行排序并将其存储到LinkedHashMap中。 LinkedHashMap保持插入顺序。
package com.journaldev.programminginterviews;import java.util.ArrayList;import java.util.HashMap;import java.util.LinkedHashMap;import java.util.List;import java.util.Map;import java.util.Map.Entry;import java.util.Set;public class SortHashMapByValue { public static void main(String[] args) { Map scores = new HashMap<>(); scores.put("David", 95); scores.put("Jane", 80); scores.put("Mary", 97); scores.put("Lisa", 78); scores.put("Dino", 65); System.out.println(scores); scores = sortByValue(scores); System.out.println(scores); } private static Map sortByValue(Map scores) { Map sortedByValue = new LinkedHashMap<>(); // get the entry set Set > entrySet = scores.entrySet(); System.out.println(entrySet); // create a list since the set is unordered List > entryList = new ArrayList<>(entrySet); System.out.println(entryList); // sort the list by value entryList.sort((x, y) -> x.getValue().compareTo(y.getValue())); System.out.println(entryList); // populate the new hash map for (Entry e : entryList) sortedByValue.put(e.getKey(), e.getValue()); return sortedByValue; }} String class doesn’t have any method to remove characters. We can use the replace() method to create a new string without the given character.
字符串类没有任何删除字符的方法。 我们可以使用replace（）方法创建一个不带给定字符的新字符串。
String str1 = "abcdABCDabcdABCD"; str1 = str1.replace("a", ""); System.out.println(str1); // bcdABCDbcdABCD The string is immutable in Java. All the string manipulation methods return a new string. So, it’s necessary that we assign it to another variable.
该字符串在Java中是不可变的。 所有的字符串操作方法都返回一个新的字符串。 因此，有必要将其分配给另一个变量。
We can create the character array from the string. Then iterate over it and create a HashMap with the character as key and their count as value.
我们可以从字符串创建字符数组。 然后对其进行迭代，并创建一个HashMap，将字符作为键，并将其计数作为值。