如何避免数组重复检查中已验证值的重复输出？

我用Utils.arrGen生成了两个指定长度、取值范围的随机数组firstArr和secondArr，用Utils类的arrRepeatCheck函数检查firstArr元素在secondArr中的出现次数并遍历输出结果，但同一值会重复打印（示例输出如下）。我试过用if判断值是否已检查但没成功，怎么解决？

原代码实现 #

数组生成代码 #

int [] firstArr = Utils.arrGen(len, min, max);
int [] secondArr = Utils.arrGen(len, min, max);

遍历检查代码 #

for (int i = 0; i < firstArr.length; i++) {
   int [] result = Utils.arrRepeatCheck(secondArr, firstArr[i]);
   if (result[1] != 0) {
      System.out.println("The number " + result[0] + " was found to be repetetive in the arrays " + result[1] + " times!");
   }
}

arrRepeatCheck函数实现 #

public static int [] arrRepeatCheck(int [] arr, int num) {
   int counter = 0;
   for (int i = 0; i < arr.length; i++) {
      if (arr[i] == num) {
         counter++;
      }
   }

   int [] result = {num, counter};
   return result;
}

注：Utils.arrGen函数用于生成指定取值范围内的随机值数组，arrRepeatCheck函数已在Utils类中实现。

当前输出结果 #

The number 1 was found to be repetetive in the arrays 12 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 1 was found to be repetetive in the arrays 12 times!
The number 1 was found to be repetetive in the arrays 12 times!
The number 2 was found to be repetetive in the arrays 8 times!
The number 0 was found to be repetetive in the arrays 15 times!
The number 1 was found to be repetetive in the arrays 12 times!
The number 0 was found to be repetetive in the arrays 15 times!
The number 2 was found to be repetetive in the arrays 8 times!
The number 0 was found to be repetetive in the arrays 15 times!
The number 2 was found to be repetetive in the arrays 8 times!
The number 1 was found to be repetetive in the arrays 12 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 2 was found to be repetetive in the arrays 8 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 1 was found to be repetetive in the arrays 12 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 0 was found to be repetetive in the arrays 15 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 0 was found to be repetetive in the arrays 15 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 2 was found to be repetetive in the arrays 8 times!
The number 1 was found to be repetetive in the arrays 12 times!
The number 3 was found to be repetetive in the arrays 15 times!
The number 0 was found to be repetetive in the arrays 15 times!
The number 3 was found to be repetetive in the arrays 15 times!

Process finished with exit code 0

解决方案 #

核心思路是记录已经检查过的元素，避免重复处理，以下是几种可行的实现方式：

方法1：用Set存储已检查的元素 #

遍历firstArr时，先判断当前元素是否已经在Set中，不在的话才执行检查并输出，同时把元素加入Set：

// 新增一个Set来记录已检查过的数字
Set<Integer> checkedNumbers = new HashSet<>();

for (int i = 0; i < firstArr.length; i++) {
    int currentNum = firstArr[i];
    // 如果该数字没检查过
    if (!checkedNumbers.contains(currentNum)) {
        int[] result = Utils.arrRepeatCheck(secondArr, currentNum);
        if (result[1] != 0) {
            System.out.println("数字 " + result[0] + " 在数组中出现了 " + result[1] + " 次！");
        }
        // 标记为已检查
        checkedNumbers.add(currentNum);
    }
}

方法2：先对firstArr去重，再遍历检查 #

先把firstArr转换成不含重复元素的集合，再遍历这个集合执行检查，从根源避免重复处理：

// 对firstArr去重
Set<Integer> uniqueFirstArr = new HashSet<>();
for (int num : firstArr) {
    uniqueFirstArr.add(num);
}

// 遍历去重后的集合
for (int num : uniqueFirstArr) {
    int[] result = Utils.arrRepeatCheck(secondArr, num);
    if (result[1] != 0) {
        System.out.println("数字 " + result[0] + " 在数组中出现了 " + result[1] + " 次！");
    }
}

方法3：优化统计逻辑，一次遍历完成统计 #

如果不想额外用集合，也可以先统计secondArr中所有元素的出现次数，再遍历firstArr的唯一元素输出，效率更高（避免多次调用arrRepeatCheck重复遍历secondArr）：

// 先统计secondArr中每个数字的出现次数
Map<Integer, Integer> countMap = new HashMap<>();
for (int num : secondArr) {
    countMap.put(num, countMap.getOrDefault(num, 0) + 1);
}

// 再遍历firstArr的唯一元素输出
Set<Integer> uniqueFirstArr = new HashSet<>();
for (int num : firstArr) {
    if (!uniqueFirstArr.contains(num)) {
        int count = countMap.getOrDefault(num, 0);
        if (count != 0) {
            System.out.println("数字 " + num + " 在数组中出现了 " + count + " 次！");
        }
        uniqueFirstArr.add(num);
    }
}

这些方法都能解决重复输出的问题，其中方法3的效率最高，因为只需要遍历secondArr一次，而原代码中每个firstArr元素都会遍历一次secondArr，数据量大时差异明显。

内容的提问来源于stack exchange，提问作者shitposter