给定一个数组，返回连续出现次数最多的元素。在Java中[关闭]

Question

在 Java 中，给定数组 A[a,a,b,b,b,c,d,e,e] 返回 'b'，因为它连续出现的次数比数组中的任何其他元素都多。我尝试将每个元素与下一个元素进行比较，如果它们匹配，则增加一个计数，然后返回计数最高的元素，但我不知道如何将它实现到代码中。

Answer 1

注意 - 这假设字符在 BMP 中

1. 保持 4 个状态 - prev character、current character、current count、max count
2. 初始prev character 为一些无效字符或为空
3. 将计数初始化为零
4. 遍历字符，如果在任何时候当前字符和前一个字符不匹配，则重置当前字符和当前计数
5. 当当前计数大于最大计数时，将当前字符更新为最大字符并更新最大计数

    public static void main(String[] args) {
        char[] input = new char[] {'a', 'b', 'b', 'b', 'b', 'c', 'd', 'e', 'e'};
        if (input.length == 0) {
            return;
        }
        char prev = (char) (input[0] - 1); // choose a character other than first character
        char maxChar = prev;
        int maxCount = 0;
        int currentCount = 0;
        for (final char current : input) {
            if (prev != current) {
                prev = current;
                currentCount = 0;
            }
            currentCount++;
            if (currentCount > maxCount) {
                maxCount = currentCount;
                maxChar = current;
            }
        }
        System.out.println(maxChar);
    }

Answer 2

只需将当前元素与前一个元素进行比较，计算连续元素的数量，如果检测到连续元素，则检查它是否大于现有最大值并跟踪该值：

public static int findMostConsecutive(int ... arr) {
    int res = arr[0];
    int count = 1;
    int maxCount = 1;
    for (int i = 1; i < arr.length; i++) {
        if (arr[i] == arr[i - 1]) {
            count++;
            if (count > maxCount) {
                maxCount = count;
                res = arr[i];
            }
        }
        else {
            count = 1;
        }
    }
    return res;
}

测试：

System.out.println("Most consecutive: " + findMostConsecutive(1, 2, 1, 1, 3, 3, 3, 4, 2));
System.out.println("Most consecutive: " + findMostConsecutive(2, 2, 2, 2, 3, 3, 3, 3));
System.out.println("Most consecutive: " + findMostConsecutive(2, 2, 2, 2, 3, 3, 3, 3, 3, 3));

输出

Most consecutive: 3
Most consecutive: 2
Most consecutive: 3