计算Java中数组中的重复次数 [英] Count repetitions in array in java

问题描述

我知道有人问过类似的问题，并且我已经进行了研究许多网站.我尝试使用一些答案，但是我的代码是仍然无法正常工作.

I know that similar questions have been asked and I have researched many websites. I have tried to use some of the answers but my code is still not working.

我正在完成以前的任务，以帮助我积累知识Java.请原谅我的代码中的任何错误，我仍在学习Java.

I am going through a previous assignment to help build my knowledge of Java. Please forgive any errors in my code, I am still learning Java.

这是我的问题:

执行一个方法计数，给定一个整数元素数组，该方法计数返回另一个数组，该数组包含输入数组中每个整数{0，...，r}的出现次数，其中r是一个整数，表示高位您需要计算的整数的边界.

Implement a method count which, given an array of integer elements, returns another array containing the number of occurrences of each integer {0, ..., r} in the input array, where r is an integer to show the upper boundary of the integers that you need to count.

返回的计数数组的大小为r + 1，其中每个索引i处的元素都对应于出现整数i的次数(其中i在{0，...，r}中).

The returned array of counts will be of size r + 1, where the element at each index i corresponds to the number of occurrences of integer i (with i in {0, ..., r}).

输入数组中从0到r的整数范围之外的元素都可以忽略.

Elements in the input array outside of the integer range from 0 to r can be ignored.

例如，假设输入[0、8、1、3、1、3、10、3]且r为4，则输出应为[1、2、0、3、0].

For example, given the input [0, 8, 1, 3, 1, 3, 10, 3] with r is 4, the output should be [1, 2, 0, 3, 0].

如果输入数组为null或长度为0，则将返回null.

If the input array is null or of length 0, this will return null.

空间要求:方法计数应仅对count数组使用额外的空间.

Space requirements: Method count should only use additional space for the count array.

时间要求:计数应通过输入数组一次计算.

Time requirements: The counts should be calculated in a single pass through the input array.

这是我到目前为止所做的事情，它不符合要求，因此我需要帮助才能找到正确的解决方案:

Here is what I've done so far, it doesn't meet the requirements so I need help in order to find the right solution:

public static int[] count(int[] arr, int r) {
        int[] count = new int[r + 1];
        for (int i = 0; i < arr.length; i++) {
            for (int j = 0; j < r; j++) {
                if (arr[i] == j) {
                    count[i]++;

                }

            }
        }
        return count;
    }

推荐答案

您真的很亲密，但似乎有些错误.

You are really close, but seems maybe a small bit is wrong.

int[] count = new int[r + 1];
for (int i = 0; i < arr.length; i++) {
    if( arr[i] <= r) {
        count[arr[i]]++;
    }
}

我认为上面的方法会起作用，如果您考虑一下，只要该索引在{0 ... r}之内，则arr的每个元素都对应于一个计数索引，因此我们检查该值是否在该索引之内范围，然后我们在count内的那个索引处增加整数.

I think the above will work, if you think about it, each element of arr corresponds to an index in count as long as that index is within {0...r}, so we check that the value is within that range, then we increment the integer at that index within count.

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