Question

我正在学习java，当我需要查找数组中每个数字的节目数时，我有一个练习，数字是0-100，这需要在O（n）中完成。现在我不知道如何做到这一点，所以我看到了解决方案，并没有真正理解它，解决方案的解释非常差，这里是代码：

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]);
    }

}

我特意没有得到这条线

temp[data[i]]+=1;

如果有人可以向我解释代码中每一行的想法，我非常感谢任何帮助。谢谢！

Answer 1

基本上，该代码执行此操作： 1）初始化具有100个位置的临时数组（全部设置为0）; 2）迭代数据数组并相对于它正在处理的数据数组值的索引递增值。举个例子：

int[] data = {23,11,10,23,5};

temp[data[0]] += 1
temp[23] += 1
temp[23] = temp[23] + 1 = 0 + 1 = 1

temp[data[1]] += 1
temp[11] += 1
temp[11] = temp[11] + 1 = 0 + 1 = 1

temp[data[2]] += 1
temp[10] += 1
temp[10] = temp[10] + 1 = 0 + 1 = 1

temp[data[3]] += 1
temp[23] += 1
temp[23] = temp[23] + 1 = 1 + 1 = 2

temp[data[4]] += 1
temp[5] += 1
temp[5] = temp[5] + 1 = 0 + 1 = 1

因此，在这种情况下，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，您可以计算O（n）中的出现次数，也就是说，只需遍历一次数组。

Answer 2

/* 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.

从我的作业中了解一些代码

2 个答案: