1024 游戲攻略｜馬爾可夫鏈，挑戰不可能的運算。

不懂游戲詳情的的可以先看鏈接

https://leetcode.cn/circle/discuss/OV9VUd/

import java.util.*; public class Main {public static List<List<Integer>> num = new ArrayList<>();public static List<List<String>> opt = new ArrayList<>();public static void main(String[] args) {// 所擁有的數字卡片int arr[] = {16, 2, 16, 0};// 所擁有的運算符卡片String brr[] = {"<<", "*", "+"};Arrays.sort(arr);Arrays.sort(brr);numdfs(new boolean[arr.length], new ArrayList<>(), arr, arr.length);optdfs(new boolean[brr.length], new ArrayList<>(), brr, brr.length);for (int i = 0; i < num.size(); i++) {List<Integer> list = num.get(i);for (int j = 0; j < opt.size(); j++) {int ans = getAns(list, opt.get(j));// 如果成功則輸出數字順序和運算符順序if (ans == 1024) {System.out.println(list);System.out.println(opt.get(j));return;}}}}public static int getAns(List<Integer> numbers, List<String> characters) {int ans = 0;for (int i = 0; i < 3; i++) {String s = characters.get(i);if (s.equals("/") && numbers.get(i + 1) == 0) return 0;if (i == 0) {ans = getAns(numbers.get(i), numbers.get(i + 1), s);} else {ans = getAns(ans, numbers.get(i + 1), s);}}return ans;}public static int getAns(int a, int b, String c) {int ans = 0;if (c.equals("+")) {ans = a + b;} else if (c.equals("-")) {ans = a - b;} else if (c.equals("*")) {ans = a * b;} else if (c.equals("//")) {ans = a / b;} else if (c.equals(">>")) {ans = a >> b;} else if (c.equals("<<")) {ans = a << b;} else if (c.equals("%")) {ans = a % b;} else if (c.equals("^")) {ans = a ^ b;} else if (c.equals("**")) {ans = (int) Math.pow(a, b);} else if (c.equals("|")) {ans = a | b;} else if (c.equals("&")) {ans = a & b;}return ans;}public static void numdfs(boolean used[], List<Integer> list, int arr[], int n) {if (list.size() == 4) {num.add(new ArrayList<>(list));return;}for (int i = 0; i < n; i++) {// 去重if (i > 0 && !used[i-1] && arr[i] == arr[i - 1]) continue;if (!used[i]) {list.add(arr[i]);used[i] = true;numdfs(used, list, arr, n);used[i] = false;list.remove(list.size() - 1);}}}public static void optdfs(boolean used[], List<String> list, String arr[], int n) {if (list.size() == 3) {opt.add(new ArrayList<>(list));return;}for (int i = 0; i < n; i++) {// 去重if (i > 0 && !used[i-1] && arr[i] == arr[i - 1]) continue;if (!used[i]) {list.add(arr[i]);used[i] = true;optdfs(used, list, arr, n);used[i] = false;list.remove(list.size() - 1);}}}

總結

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