阿里笔试2023-3-15

太菜了，记录一下笔试题目，代码有更好解法欢迎分享。
1、满二叉子树的数量。
给定一颗二叉树，试求这课二叉树有多少个节点满足以该节点为根的子树是满二叉树？满二叉树指每一层都达到节点最大值。
第一行输入n表示节点数量，接下来n行第一个代表左儿子，第二个代表右儿子。

public class Main {static int res = 0;public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();int[][] nums = new int[n][2];for (int i = 0; i < n; i ++) {nums[i][0] = in.nextInt();nums[i][1] = in.nextInt();}new Main().isFullTree(nums, 1)；System.out.println(res);}public int height(int[][] nums, int root) {if (root == -1) {return 0;}return Math.max(height(nums, nums[root - 1][0]), height(nums, nums[root - 1][1])) + 1;}public boolean isFullTree(int[][] nums, int root) {if (root == -1) {return true;}if (isFullTree(nums, nums[root - 1][0]) && isFullTree(nums, nums[root - 1][1]) && height(nums, nums[root - 1][0]) == height(nums, nums[root - 1][1])) {res ++;return true;} else {return false;}}
}

上述解法时间复杂度O(n)，空间复杂度o()
另外，可以掌握根据数组进行二叉树建树

class TreeNode {TreeNode left;TreeNode right;int val;public TreeNode() {}public TreeNode(int val) {this.val = val;}public TreeNode (TreeNode left, TreeNode right, int val) {this.left = left;this.right = right;this.val = val;}
}public static void main(String[] args) {List<TreeNode> nodeList = new ArrayList<>();nodeList.add(null);for (int i = 1; i <= n; i ++) {nodeList.add(new TreeNode(i));}for (int i = 0; i < n; i ++) {if (nums[i][0] == -1) {nodeList.get(i + 1).left = null;} else {nodeList.get(i + 1).left = nodeList.get(nums[i][0]);}if (nums[i][1] == -1) {nodeList.get(i + 1).right = null;} else {nodeList.get(i + 1).right = nodeList.get(nums[i][1]);}}
}

2、三元组计数。
给定一个数组，计算有多少个三元组0<=i<j<k<n，且max(nums[i], nums[j], nums[k]) - min(nums[i], nums[j], nums[k]) = 1。
第一行输入n表示数组个数，第二行输入n个整数。

public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int[] nums = new int[n];for (int i = 0; i < n; i++) {nums[i] = scanner.nextInt();}Arrays.sort(nums);Map<Integer, Integer> map = new HashMap<>();int res = 0;for (int i = 0; i < n; i ++) {map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);}for (int i = 0; i < n; i ++) {if (i > 0 && nums[i] == nums[i - 1]) {continue;}if (map.containsKey(nums[i] + 1)) {int low = map.get(nums[i]);int high = map.get(nums[i] + 1);if (low == 1 && high == 1) {continue;}if (high > 1) {res += low * high * (high - 1) / 2;}if (low > 1) {res += high * low * (low - 1) / 2;}}}System.out.println(res);
}

上述解法时间复杂度O(nlog(n))，空间复杂度O(n)。
3、乘2除2。
在n个元素的数组中选择k个元素，每个元素要么乘以2，要么除以2并向下取整，使得操作完后数组的极差尽可能小，并且输出极差。极差为最大值减去最小值。
第一行输入整数n和k。第二行输入n个整数表示数组。

public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int k = scanner.nextInt();int[] nums = new int[n];for (int i = 0; i < n; i++) {nums[i] = scanner.nextInt();}Comparator<Integer> comparator = new Comparator<Integer>() {@Overridepublic int compare(Integer o1, Integer o2) {return o2 - o1;}};Arrays.sort(nums);PriorityQueue<Integer> queueMin = new PriorityQueue<>(comparator);PriorityQueue<Integer> queueMid = new PriorityQueue<>(comparator);PriorityQueue<Integer> queueMax = new PriorityQueue<>(comparator);int minMin = Integer.MAX_VALUE, midMin = Integer.MAX_VALUE, maxMin = Integer.MAX_VALUE;for (int i = 0; i < k ; i ++) {minMin = Math.min(minMin, 2 * nums[i]);queueMin.add(2 * nums[i]);}for (int i = k; i < n; i ++) {midMin = Math.min(midMin, nums[i]);queueMid.add(nums[i]);}int res = Integer.MAX_VALUE;if (k == 0) {res = Math.min(res, queueMid.peek() - midMin);} else {res = Math.min(res, Math.max(queueMin.peek(), queueMid.peek()) - Math.min(minMin, midMin));}for (int i = 0; i < k; i ++) {int tempMin = queueMin.poll();queueMid.add(tempMin / 2);midMin = Math.min(midMin, tempMin / 2);int tempMid = queueMid.poll();queueMax.add(tempMid / 2);maxMin = Math.min(maxMin, tempMid / 2);if (i == k - 1) {res = Math.min(res, Math.max(queueMid.peek(), queueMax.peek()) - Math.min(Math.min(minMin, midMin), maxMin));} else {res = Math.min(res, Math.max(Math.max(queueMin.peek(), queueMid.peek()), queueMax.peek()) - Math.min(Math.min(minMin, midMin), maxMin));}}System.out.println(res);
}

上述解法时间复杂度O(nlog(n))，空间复杂度O(n)。