java - 二维数组中最常见的元素。如何优化？

我有这个任务，找到int [] []数组中最常见的元素，并打印重复的次数和次数。我解决了这个问题。

我的一个朋友说拥有4个for()是个坏主意，所以我决定尝试对其进行优化以删除一两个，但找不到办法。

所以这里是：

     int cnt, element = arr[0][0], numberRepeats = 0;//cnt-counter,what's the element ,how many times it's repeated
     for (int i = 0; i < arr.length; i++) {
        for (int j = 0; j < arr[i].length; j++) {//those two for's are for the current element
            cnt = 0;//counter is nullified
            for (int j2 = i; j2 < arr.length; j2++) {
                for (int k = 0; k < arr[j2].length; k++) {//and those two are the compared element
                    if (arr[i][j] == arr[j2][k]) {//if the current element is the same as the compared element,increase counter
                        cnt++;
                    }
                }

            if (cnt > numberRepeats) {//after the compared element is done comparing and the number of repeats of the current element is more then the lastly checked element
                element = arr[i][j];//we get the element ,and how many times it's repeated
                numberRepeats = cnt;
            }
        }
    }
}

我能想到的唯一优化是从当前元素开始计数直到结束。

请给我更多的想法。我能找到的所有答案都是关于一维数组的，它们几乎都是相同的代码。

最佳答案

您可以迭代整个数组并计算映射中每个元素的出现次数。

Map<Integer, Integer> elementsCounts = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
    for (int j = 0; j < arr[i].length; j++) {
        Integer count = elementsCounts.get(arr[i][j]])
        if(count == null){
          count = 0
        }
        elementsCounts.put(arr[i][j]], count+1)
    }
}

现在，获得的只是在地图中找到具有最大值的键。