> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/graph-search/679.24-game.md).

# 679.24-Game

## 679. 24 Game

## 题目地址

<https://leetcode.com/problems/24-game/>

## 题目描述

```
You have 4 cards each containing a number from 1 to 9. You need to judge whether they could operated through *, /, +, -, (, ) to get the value of 24.

Example 1:
Input: [4, 1, 8, 7]
Output: True
Explanation: (8-4) * (7-1) = 24

Example 2:
Input: [1, 2, 1, 2]
Output: False

Note:
The division operator / represents real division, not integer division. For example, 4 / (1 - 2/3) = 12.
Every operation done is between two numbers. In particular, we cannot use - as a unary operator. For example, with [1, 1, 1, 1] as input, the expression -1 - 1 - 1 - 1 is not allowed.
You cannot concatenate numbers together. For example, if the input is [1, 2, 1, 2], we cannot write this as 12 + 12.
```

## 代码

### Approach 1: backtracing

```java
class Solution {
    public boolean judgePoint24(int[] nums) {
        ArrayList A = new ArrayList<Double>();
        for (int v: nums) A.add((double) v);
        return solve(A);
    }

    private boolean solve(ArrayList<Double> nums) {
        if (nums.size() == 0) return false;
        if (nums.size() == 1) return Math.abs(nums.get(0) - 24) < 1e-6;

        for (int i = 0; i < nums.size(); i++) {
            for (int j = 0; j < nums.size(); j++) {
                if (i != j) {
                    ArrayList<Double> nums2 = new ArrayList<Double>();
                    for (int k = 0; k < nums.size(); k++) {
                      if (k != i && k != j) {
                        nums2.add(nums.get(k));
                        }
                    }
                    for (int k = 0; k < 4; k++) {
                        if (k < 2 && i < j) continue; // 避免加法乘法的交换性
                        if (k == 0) nums2.add(nums.get(i) + nums.get(j));
                        if (k == 1) nums2.add(nums.get(i) * nums.get(j));
                        if (k == 2) nums2.add(nums.get(i) - nums.get(j));
                        if (k == 3) {
                            if (nums.get(j) != 0) {
                                nums2.add(nums.get(i) / nums.get(j));
                            } else {
                                continue;
                            }
                        }
                        if (solve(nums2)) return true;
                        nums2.remove(nums2.size() - 1);
                    }
                }
            }
        }
        return false;
    }
}
```

### Backtracking

```java
class Solution {
    public boolean judgePoint24(int[] nums) {
        List<Double> list = new ArrayList<>();
        for (int i : nums) {
            list.add((double) i);
        }
        return dfs(list);
    }

    // 每次dfs都是选取两张牌
    private boolean dfs(List<Double> list) {
        if (list.size() == 1) {
            // 如果此时list只剩下了一张牌
            if (Math.abs(list.get(0) - 24.0) < 0.001) {
                return true;
            }
            return false;
        }

        // 选取两张牌
        for (int i = 0; i < list.size(); i++) {
            for (int j = i + 1; j < list.size(); j++) {
                // 对于每下一个可能的产生的组合
                for (double c : compute(list.get(i), list.get(j))) {
                    List<Double> nextRound = new ArrayList<>();
                    // 将他们加入到下一个list循环中去
                    nextRound.add(c);
                    for (int k = 0; k < list.size(); k++) {
                        if (k == j || k == i)  continue;
                        nextRound.add(list.get(k));
                    }
                    if (dfs(nextRound)) return true;
                }
            }
        }
        return false;
    }

    // 计算下一个可能产生的组合
    private List<Double> compute(double a, double b) {
        List<Double> res = Arrays.asList(a + b, a - b, b - a, a * b, a / b, b / a);
        return res;
    }
}
```


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