我目前正在制作一个游戏,我正在尝试比较屏幕上显示的最短路径(这是随机生成的整数路径(和用户采用的路径。这部分代码是我尝试获取屏幕上总路径Math.min()
的代码:
int path1 = randomNum3 + randomNum8;
int path2 = randomNum2 + randomNum7;
int path3 = randomNum + randomNum6;
int path4 = randomNum3 + randomNum5 + randomNum7;
int path5 = randomNum2 + randomNum5 + randomNum8;
int path6 = randomNum + randomNum4 + randomNum7;
int path7 = randomNum2 + randomNum4 + randomNum6;
int caminoMasCorto = Math.min(path1, path2, path3, path4, path5,
path6, path7);
向我显示的错误是这样的:
找不到适合 min(
int,int,int,int,int,int,int,int,int( 的方法方法Math.min(int,int(不适用
(实际参数列表和正式参数列表的长度不同(
要找到许多int
变量的最小值,请使用:
int caminoMasCorto = IntStream.of(path1, path2, path3, path4, path5,
path6, path7).min().getAsInt();
如果您没有使用 Java 8+,则可以执行以下操作:
int caminoMasCorto = path1;
for (int value : new int[] { path2, path3, path4, path5, path6, path7 })
caminoMasCorto = Math.min(caminoMasCorto, value);
是的。正如错误所说,Math.min(int, int)
需要两个参数。改变
int caminoMasCorto = Math.min(path1, path2, path3, path4, path5,
path6, path7);
自
int caminoMasCorto = Math.min(path1, Math.min(path2,
Math.min(path3, Math.min(path4, Math.min(path5, Math.min(path6, path7))))));
或(如安德烈亚斯所指出的(
int caminoMasCorto = Math.min(Math.min(Math.min(Math.min(Math.min(Math.min(
path1, path2), path3), path4), path5), path6), path7);
或者,您可以编写自己的min
函数,该函数采用任意数量的int
并返回最小值。喜欢
private static int myMin(int... vals) {
if (vals == null || vals.length < 1) {
throw new RuntimeException("No values");
}
int t = vals[0];
for (int i = 1; i < vals.length; i++) {
t = Math.min(t, vals[i]);
}
return t;
}
然后使用
int caminoMasCorto = myMin(path1, path2, path3, path4, path5,
path6, path7);
使用您认为最具可读性的任何一种。
Java的Math.min
接受两个参数。要将其扩展到多个数字,请尝试此方法:
public int minExtension (int... numbers) {
int minimum = numbers[0];
for (int number : numbers) {
minimum = Math.min(number, minimum)
}
return minimum
}
这样称呼它
int caminoMasCorto = minExtension (path1, path2, path3, path4, path5, path6, path7);