每次生成随机数，不包含最后一个数字 [英] Generate random number each time and don't include last number

问题描述

我有4种颜色.我要这样做，以使播放器不能连续两次出现相同的颜色.当玩家与某个对象碰撞时，会调用RandomColor().因此该功能在游戏中会被多次调用，有时玩家不会改变其颜色.

I have 4 colors. I want to make it so that the player can't be the same color 2 times in a row. When a player collides with an object, the RandomColor() is called. So this function is called many times during the game and sometimes the player does not change his color.

 using UnityEngine;

 public class ColorManager : MonoBehaviour {

     public SpriteRenderer player;
     public string playerColor;

     public Color[] colors = new Color[4];


     private void Awake()
     {
         RandomColor();        
     }

     public void RandomColor()
     {
         int index = Random.Range(0, 4);

         switch (index)
         {
             case 0:
                 player.color = colors[0]; //colors[0] is orange
                 playerColor = "orange";
                 break;

             case 1:
                 player.color = colors[1]; //colors[1] is pink
                 playerColor = "pink";
                 break;

             case 2:
                 player.color = colors[2]; //colors[2] is blue
                 playerColor = "blue";
                 break;

             case 3:
                 player.color = colors[3]; //colors[3] is purple
                 playerColor = "purple";
                 break;
             }    
     }    
 }

尝试使用while循环，执行while循环，但是显然我做错了，因为有时我连续两次收到相同的颜色.谁能弄清楚并解释其工作方式/原因，那将是很棒的，因为我在这个问题上花了很多时间，而且我很好奇.

Tried using while loop, do while loop, but I'm obviously doing it wrong, since I receive the same color twice in a row sometimes. It would be great if anyone figures it out and explains how/why it works, because I spent a big chunk of time on the issue and I am very curious.

推荐答案

首先，您需要一个函数，该函数可以生成排除后的随机数.以下是我为此使用的内容:

First, you need a function that can generate a random number with exclusion. Below is what I use for that:

int RandomWithExclusion(int min, int max, int exclusion)
{
    int result = UnityEngine.Random.Range(min, max - 1);
    return (result < exclusion) ? result : result + 1;
}

每次调用它时，都需要将结果存储在一个全局变量中，以便在下次再次调用它时将其传递给exclusion参数.

Each time you call it, you need to store the result in a global variable so that you will pass that to the exclusion parameter next time you call it again.

我修改了该函数，以便您不必在每次调用该函数时都这样做.新的RandomWithExclusion函数将为您做到这一点.

I modified the function so that you don't have to do that each time it is called. The new RandomWithExclusion function will do that for you.

int excludeLastRandNum;
bool firstRun = true;

int RandomWithExclusion(int min, int max)
{
    int result;
    //Don't exclude if this is first run.
    if (firstRun)
    {
        //Generate normal random number
        result = UnityEngine.Random.Range(min, max);
        excludeLastRandNum = result;
        firstRun = false;
        return result;
    }

    //Not first run, exclude last random number with -1 on the max
    result = UnityEngine.Random.Range(min, max - 1);
    //Apply +1 to the result to cancel out that -1 depending on the if statement
    result = (result < excludeLastRandNum) ? result : result + 1;
    excludeLastRandNum = result;
    return result;
}

测试:

void Update()
{
    Debug.Log(RandomWithExclusion(0, 4));
}

最后一个数字将永远不会出现在下一个函数调用中.

The last number will never appear in the next function call.

对于您的特定解决方案，只需替换

For your specific solution, simply replace

int index = Random.Range(0, 4);

使用

int index = RandomWithExclusion(0, 4);

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