Java 理解作业中的一些代码

我正在学习java，我有一个练习，当我需要找到数组中每个数字的显示次数时，数字是0-100，这需要在O（n）中完成。现在我不知道怎么做，所以我看到了解决方案，并没有真正理解它，对解决方案的解释很差，下面是代码：

public static void count (int [] data){
    final int N = 100;
    int [] temp = new int[N];
    int i;
    for (i=0; i < data.length; i++){
        temp[data[i]]+=1;  
    }
    for (i=0; i < N; i++){
        if ( temp[i] != 0)
            System.out.println(i + ":" +temp[i]);
    }

}

如果有人能解释代码中每一行的意思，

---

我真的很感谢你的帮助。谢谢

基本上，该代码是这样做的： 1） 用100个位置初始化临时数组（全部设置为0）； 2） 迭代数据数组，并相对于正在处理的数据数组值的索引增加值。举个例子：

int[]data={23,11,10,23,5}；
临时[数据[0]]+=1
温度[23]+=1
温度[23]=温度[23]+1=0+1=1
温度[数据[1]]+=1
温度[11]+=1
温度[11]=温度[11]+1=0+1=1
温度[数据[2]]+=1
温度[10]+=1
温度[10]=温度[10]+1=0+1=1
温度[数据[3]]+=1
温度[23]+=1
温度[23]=温度[23]+1=1+1=2
温度[数据[4]]+=1
温度[5]+=1
温度[5]=温度[5]+1=0+1=1

在本例中，您会注意到值23在数据数组中出现两次，而其他值只出现一次。这样，您就可以计算O（n）中的出现次数，也就是说，只需遍历一次数组。

/*方法签名
/* method signature
static - method can be used without instantiation of the class
public - can be accessed by other classes
void - the return of the function count.
int [] data - parameter of the function. This function receives an array of int
*/

public static void count (int [] data){
    final int N = 100;
    int [] temp = new int[N];
    int i;

 //for i = 0 until i < size of the array passed as parameter, increment 1 in i

    for (i=0; i < data.length; i++){

/*
each iteration of this for increments 1 in the array temp at the position informed by the value of parameter data in position i.
caution: this can bring serious problems of segmentation fault, in other words, you can try to store in a position of the array temp that doesn't exist.
This will happen because you cannot guarantee the number that will be store in data[i]. If data[i] > 99 (because N = 100), you will face a segmentation fault.
*/

        temp[data[i]]+=1;  
    }

//here you will print the value stored at temp array in position i. From 0 to 99   

    for (i=0; i < N; i++)
    {
            if ( temp[i] != 0)
                System.out.println(i + ":" +temp[i]);
    }

}

HOPE IT HELPS.

You need to study more. Your question is so naive.

静态-方法可以在不实例化类的情况下使用
公共-可由其他类访问
void-返回函数计数。
int[]data—函数的参数。此函数接收一个int数组
*/
公共静态无效计数（int[]数据）{
最终整数N=100；
int[]temp=新的int[N]；
int i；
//对于i=0，直到i<作为参数传递的数组的大小，在i中增加1
对于（i=0；i99（因为N=100），您将面临分段错误。
*/
温度[数据[i]]+=1；
}
//在这里，您将打印存储在位置i的临时数组中的值，从0到99
对于（i=0；i

解释是什么？你从中不明白什么？有道理。非常感谢。

/* method signature
static - method can be used without instantiation of the class
public - can be accessed by other classes
void - the return of the function count.
int [] data - parameter of the function. This function receives an array of int
*/

public static void count (int [] data){
    final int N = 100;
    int [] temp = new int[N];
    int i;

 //for i = 0 until i < size of the array passed as parameter, increment 1 in i

    for (i=0; i < data.length; i++){

/*
each iteration of this for increments 1 in the array temp at the position informed by the value of parameter data in position i.
caution: this can bring serious problems of segmentation fault, in other words, you can try to store in a position of the array temp that doesn't exist.
This will happen because you cannot guarantee the number that will be store in data[i]. If data[i] > 99 (because N = 100), you will face a segmentation fault.
*/

        temp[data[i]]+=1;  
    }

//here you will print the value stored at temp array in position i. From 0 to 99   

    for (i=0; i < N; i++)
    {
            if ( temp[i] != 0)
                System.out.println(i + ":" +temp[i]);
    }

}

HOPE IT HELPS.

You need to study more. Your question is so naive.