扑克牌魔术里面的算法

布满荆棘的人生 2023-01-03 14:30 177阅读 0赞

# 前言 #

今天在 B 站上看了一个和扑克牌、算法、数学有关的心灵感应魔术，适合撩妹子。

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https://b23.tv/uXIQeL

看完之后，想写一篇和扑克牌有关的算法文章。

# 正文 #

## 题目描述 ##

从扑克牌中随机抽 5 张牌，判断是不是一个顺子，即这 5 张牌是不是连续的。2～10 为数字本身，A 为 1 ，J 为 11 ，Q 为 12 ，K 为 13 ，而大、小王为 0 ，可以看成任意数字。A 不能视为 14 。

**示例 1:**

输入: [1,2,3,4,5]
    输出: True

**示例 2:**

输入: [0,0,1,2,5]
    输出: True

**限制：**

*  数组长度为 5
 *  数组的数取值为 \[0, 13\]

## 题目分析 ##

这个题目还是挺简单的。

直接遍历看看是否有重复的牌，有的话直接返回 false，否则找出这五张牌最大值和最小值，由于大、小王都为 0 且可以充当任意数字，所以要先 0 不在寻找最大值和最小值的考虑范围内，最后使用最大值减去最小值，判断结果是否小于 5 ， 如果小于 5 说明加上可以看出任意数的大小王之后可以构成顺子，否则不行。

### 动画理解如下： ###

### 图片： ###

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<<< 左右滑动见更多 >>>

### 参考代码： ###

class Solution {
        public boolean isStraight(int[] nums) {
            HashSet<Integer> set = new HashSet<>();
            //数组的数取值为 [0, 13] 
            //所以可以设置最大值为 0 ，最小值为 14，这样在一开始遍历的过程中出现非 0 的情况时，总是会比最大值 0 更大，比最小值 14 更新，从而更新最大最小值
            int max = 0, min = 14;
            
            for(int num : nums){
                //跳过大小王
                if(num == 0){
                     continue;
                }
                //记录最大最小值
                min = Math.min(num, min);
                max = Math.max(num, max);
                 // 如果存在重复元素，直接返回false
                if(set.contains(num)){
                    return false;         
                }
                //记录符合要求的牌
                set.add(num);
            }
            //判断最大最小值之差是否小于 5
            return max - min < 5;
        }
    }

--------------------

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如果你有想交流的话题，欢迎留言。

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