Mastering data structures and algorithms is an essential skill for Java development interviews. This article analyzes common Java data structures (arrays, linked lists, stacks, queues) and algorithms (sorting algorithms, search algorithms), and provides a practical case: finding two numbers from an array whose sum is a specified target value.
Java Data Structures and Algorithms: Practical Analysis of Interviews
Mastering data structures and algorithms is a must for Java developers during interviews Skill. This article provides an in-depth analysis of common data structures and algorithms in Java through clear and simple language, and provides practical examples to help you stand out in interviews.
Data structure
Array: An ordered collection of elements, accessed using index. Elements can be accessed randomly in O(1) time.
int[] arr = new int[10]; arr[0] = 5; System.out.println(arr[0]); // 输出:5
Linked list: A collection of nodes, each node contains data and a reference to the next node. Provides O(1) insertions and deletions, but lookups take O(n) time.
LinkedList<String> list = new LinkedList<>(); list.add("元素 1"); list.add("元素 2"); System.out.println(list.get(0)); // 输出:元素 1
Stack: A data structure that follows the last-in-first-out (LIFO) principle. The push() and pop() operations add and remove elements in O(1) time respectively.
Stack<Integer> stack = new Stack<>(); stack.push(10); stack.push(20); System.out.println(stack.pop()); // 输出:20
Queue: Follows the first-in-first-out (FIFO) principle. The enqueue() and dequeue() operations add and remove elements in O(1) time respectively.
Queue<String> queue = new LinkedList<>(); queue.add("元素 1"); queue.add("元素 2"); System.out.println(queue.remove()); // 输出:元素 1
Algorithm
## Sorting algorithm:
Search algorithm:
Practical case:
Scenario: Find two values whose sum is the specified target value from the array number.
Code:
public static int[] findTwoSum(int[] nums, int target) { for (int i = 0; i < nums.length; i++) { for (int j = i + 1; j < nums.length; j++) { int sum = nums[i] + nums[j]; if (sum == target) { return new int[] {i, j}; } } } return null; // 未找到 }
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