在剪刀石头布游戏中确定获胜者 [英] determining the winner in a rock paper scissor game

问题描述

在这款剪刀石头布游戏中，我需要一种不同的方法来确定获胜者.当我运行程序时，其余程序运行平稳，但我不会显示获胜者.我认为它们可能是我代码中的逻辑错误.

in this rock paper scissor game i need an different method to determine winner. when i run my program the rest runs smoothly but i wont display a winner. i think their might be a logical error in my code.

public static void determineWinner(String userLower, String comp) {

      String user = userLower;
      String computer = comp;

      if(user.equals(computer)) {
            System.out.println("It's a tie!");
      } else if(user.equals("rock")) {
            if(computer.equals("scissors")) {
                  System.out.println("You win!");
            }
      } else if(computer.equals("rock")) {
            if(user.equals("scissors")) {
                  System.out.println("You lose!");
            }
      } else if(user.equals("scissors")) {
            if(computer.equals("paper")){
                  System.out.println("You win!");
            }
      } else if(computer.equals("scissors")) {
            if(user.equals("paper")) {
                  System.out.println("You lose");
            }
      } else if(user.equals("paper")) {
            if(computer.equals("rock")) {
                  System.out.println("You Win!");
            }
      } else if(computer.equals("paper")) {
            if(user.equals("rock")) {
                  System.out.println("You lose!");
            }    
      }       

推荐答案

基本上，您正在测试每种可能的组合(除了平局).使用剪刀石头布，您会有9种不同的结果，其中3种是平局，所以还剩下6种.

Basically, you are testing every possible combination (with the exception of a tie). With rock-paper-scissors, you have 9 different outcomes, and 3 of them are ties so there are 6 more left.

现在想像一下，您必须针对21世纪的变体石头剪刀布-蜥蜴-Spock"​​执行此操作:25种不同的结果.这是if else的一长串.

Now imagine you have to do this for the 21st century variant "rock-paper-scissors-lizard-Spock": 25 different outcomes. That's a long list of if elses.

但是:所有可能的结果都可以存储在一个表中，并且该表可以随您想要的大小而变.该表用于RPSLS(找不到用于RPS的表)，但是原理是相同的:

However: all possible outcomes can be stored into a table, and that table can be as small (or large) as you want. This table is for RPSLS (can't find one for RPS), but the principle is the same:

您在垂直方向选择您的选择，在水平方向选择您的计算机；那么，左边的数字就是对您的得分(显然，Win代表1，Lose代表输)，右边的数字是对手.

You pick your choice vertically and the computer's horizontally; then, the number on the left is points for you (obviously, 1 is for Win, -1 is Lose) and the number on the right is the opponent.

(编辑)这是剪刀石头布的表格.每个组合不需要两个元素，另一个总是与显示的相反.该表在列中显示了获胜者，在行中显示了失败者.

(Edit) Here is the table for Rock-Paper-Scissors. You don't need two elements per combination, the other one is always the opposite of what's shown. The table shows Winner in the columns, Loser in the rows.

+----------+-------+-------+--------+
|          | Rock  | Paper |Scissors|
+----------+-------+-------+--------+
| Rock     |   0   |   1   |   -1   |
| Paper    |  -1   |   0   |    1   |
| Scissors |   1   |  -1   |    0   |
+----------+-------+-------+--------+

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