设计类

使用Redis实现一个用户一分钟抽奖3次写一个你觉得最好的单例模式，你觉得这么写还有什么问题？ 单例模式的几种写法 饿汉式懒汉式、双重检查锁静态内部类 public class Singleton { private Singleton(){} private static class SingletonHolder { private static final Singleton INSTANCE = new Singleton(); } public static Singleton getInstance() { return SingletonHolder.INSTANCE; } }

枚举类

public enum SingletonEnum { INSTANCE(1, "aa"); SingletonEnum(int value, String v) { } } - 单例模式怎么破坏 策略模式demo 我的github上的demo工厂模式demo 设计模式demo写一个观察者模式demo demo模板方法demo

算法题

排序 冒泡排序归并排序选择排序插入排序快速排序 public int[] sortArray(int[] nums) { if (nums == null) { return nums; } quickSort(nums, 0, nums.length - 1); return nums; } public void quickSort(int[] numbers, int left, int right) { if (right <= left) { return; } int pivot = numbers[left]; int start = left; int end = right; while (start < end) { while (start < end && pivot <= numbers[end]) { end--; } swapArr(numbers, start, end); while (start < end && numbers[start] <= pivot) { start++; } swapArr(numbers, start, end); } quickSort(numbers, left, start - 1); quickSort(numbers, start + 1, right); } private void swapArr(int[] numbers, int left, int right) { int temp = numbers[left]; numbers[left] = numbers[right]; numbers[right] = temp; } 树的前中后序遍历 /*深度优先*/ /** * 先序遍历 * @param root * @param linkedList */ public static void preOrder(TreeNode root, List<TreeNode> linkedList) { if (root != null) { linkedList.add(root); preOrder(root.left, linkedList); preOrder(root.right, linkedList); } } /** * 中序遍历 * @param root * @param linkedList */ public static void inOrder(TreeNode root, List<TreeNode> linkedList) { if (root != null) { inOrder(root.left, linkedList); linkedList.add(root); inOrder(root.right, linkedList); } } /** * 后序遍历 * @param root * @param linkedList */ public static void postOrder(TreeNode root, List<TreeNode> linkedList) { if (root != null) { postOrder(root.left, linkedList); postOrder(root.right, linkedList); linkedList.add(root); } } 给前序中序遍历的结果还原二叉树斐波拉契数列二分查找 public int search(int[] nums, int target) { if (nums == null) { return -1; } int low = 0; int high = nums.length - 1; while (low <= high) { int middle = (low + high)/2; if (nums[middle] == target) { return middle; } else if (nums[middle] < target) { low = middle + 1; } else { high = middle - 1; } } return -1; } 单链表翻转 public ListNode reverseList(ListNode head) { ListNode pre = null; ListNode curr = head; while (curr != null) { ListNode next = curr.next; curr.next = pre; pre = curr; curr = next; } return pre; } 数组中数字出现的次数两数之和 2sum public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { int temp = target - nums[i]; if (map.containsKey(temp)) { return new int[]{map.get(temp), i}; } map.put(nums[i], i); } return null; } 判断链表是否有环 // 基础版 进阶版可以使用快慢指针 public ListNode detectCycle(ListNode head) { Set<ListNode> nodeSet = new HashSet<ListNode>(); ListNode item = head; int index = 0; while (item != null) { if (nodeSet.contains(item)) { return item; } nodeSet.add(item); item = item.next; } return null; } 合并两个有序数组 // 双指针 public void merge(int[] nums1, int m, int[] nums2, int n) { int[] temp = new int[nums1.length]; int mp = 0, np = 0; for (int i = 0; i < temp.length; i++) { if (mp < m && np < n) { if (nums1[mp] < nums2[np]) { temp[i] = nums1[mp++]; } else if (nums2[np] < nums1[mp]) { temp[i] = nums2[np++]; } else { temp[i++] = nums1[mp++]; temp[i] = nums2[np++]; } continue; } if (np < n) { temp[i] = nums2[np++]; continue; } if (mp < m) { temp[i] = nums1[mp++]; } } if (m >= 0) System.arraycopy(temp, 0, nums1, 0 ,temp.length); } 合并两个数组 可以先排序，后合并回文数 先翻转，再判断翻转整数 public int reverse(int x) { long result = 0; while (x != 0) { result = result*10 + x%10; x = x/10; } // 注意判断越界 return result > 0x7fffffff || result < 0x80000000 ? 0 : (int)result; } 删除链表倒数第n个元素 public ListNode removeNthFromEnd(ListNode head, int n) { int index = - n; ListNode item = head; while (item != null) { item = item.next; index++; } ListNode pre = null; ListNode curr = head; while (curr != null) { if (index-- == 0) { if (pre == null) { return curr.next; } pre.next = curr.next; break; } pre = curr; curr = curr.next; } return head; } 正方形，从左上角到右下角总共条路两个有序数组，第n大的数一个长度n的数组，前k（k<n）个升序，后面的降序求二叉树的最大深度