Java & TypeScript 游戏等级匹配算法：简单到难的经验等级匹配方法 (0-100 级)

以下是一个使用Java和Typescript设计游戏经验等级匹配方法的示例代码：\n\nJava：\n\njava\npublic class GameLevelMatcher {\n public static void main(String[] args) {\n int playerLevel = 75;\n int matchedLevel = matchLevel(playerLevel);\n System.out.println("匹配到的游戏等级为: " + matchedLevel);\n }\n\n public static int matchLevel(int playerLevel) {\n if (playerLevel < 0 || playerLevel > 100) {\n throw new IllegalArgumentException("玩家等级必须在0-100之间");\n }\n\n if (playerLevel <= 10) {\n return 10;\n } else if (playerLevel <= 20) {\n return 20;\n } else if (playerLevel <= 30) {\n return 30;\n } else if (playerLevel <= 40) {\n return 40;\n } else if (playerLevel <= 50) {\n return 50;\n } else if (playerLevel <= 60) {\n return 60;\n } else if (playerLevel <= 70) {\n return 70;\n } else if (playerLevel <= 80) {\n return 80;\n } else if (playerLevel <= 90) {\n return 90;\n } else {\n return 100;\n }\n }\n}\n\n\nTypescript：\n\ntypescript\nfunction matchLevel(playerLevel: number): number {\n if (playerLevel < 0 || playerLevel > 100) {\n throw new Error("玩家等级必须在0-100之间");\n }\n\n if (playerLevel <= 10) {\n return 10;\n } else if (playerLevel <= 20) {\n return 20;\n } else if (playerLevel <= 30) {\n return 30;\n } else if (playerLevel <= 40) {\n return 40;\n } else if (playerLevel <= 50) {\n return 50;\n } else if (playerLevel <= 60) {\n return 60;\n } else if (playerLevel <= 70) {\n return 70;\n } else if (playerLevel <= 80) {\n return 80;\n } else if (playerLevel <= 90) {\n return 90;\n } else {\n return 100;\n }\n}\n\nconst playerLevel: number = 75;\nconst matchedLevel: number = matchLevel(playerLevel);\nconsole.log("匹配到的游戏等级为: " + matchedLevel);\n\n\n这个例子中，我们使用一个简单的分段逻辑来匹配玩家的游戏等级。根据玩家的等级范围，将其分配到对应的游戏等级。如果玩家的等级不在0-100之间，将抛出一个异常（Java）或错误（Typescript）。\n\n在这个例子中，玩家等级0-10将匹配到10级，11-20将匹配到20级，以此类推，90-100将匹配到100级。