C#-求int数组中连续偶数列的个数

Posted 2020-11-17 zhyue93

求int数组中连续偶数列的个数

例如：[3, 3, 2, 2, 2, 4, 3, 5, 4, 6, 3]=>2,2,2,4;4,6 结果为2

　　    [3, 3, 2,3, 2, 2, 4, 3, 5, 4, 6, 3]=>2;2,2,4;4,6 结果为3

实现思路：

将数组取余转换为01数组[3, 3, 2, 2, 2, 4, 3, 5, 4, 6, 3]=>[1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1]

算出连续为0的个数就是偶数列的个数

int[] arr = { 3, 3, 2, 2, 2, 4, 3, 5, 4, 6, 3 };
int[] arrNew = new int[arr.Length];
for (int i = 0; i < arr.Length; i++)
{
    Console.Write(arr[i]);
    int result = arr[i] % 2;
    arrNew[i] = result;
}
Console.WriteLine();
int oddCount = 0, evenCount = 0;
int current = 0, next = 0;
for (int i = 0; i < arrNew.Length - 1; i++)
{
    current = arrNew[i];
    Console.Write(current);
    next = arrNew[i + 1];
    if (current == 0 && next != current)
        evenCount++;
    if (current == 1 && next != current)
        oddCount++;
    if (i == arrNew.Length - 2)
    {
        if (next == 0) evenCount++;
        else oddCount++;
    }
}
Console.WriteLine(arrNew[arrNew.Length - 1]);
Console.WriteLine("奇数：{0}，偶数：{1}", oddCount, evenCount);
Console.ReadKey();