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Program.cs
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Program.cs
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//创建字典record,计算以每个数字为结尾,向前数奇数的个数,将这个个数的出现的频率记入record。相当于记录了在几个位置上,之前有该个数的奇数
//遍历record,如果发现当前奇数个数减去k也在record中,则证明这两个出现频率之间的所有子串都满足条件。用当前计数个数出现的频率乘以kv.key - k上的频率,并计入res。
using System;
using System.Collections.Generic;
namespace CountNumberOfNiceSubarrays
{
class Program
{
static void Main(string[] args)
{
int[] nums = { 2, 2, 2, 1, 2, 2, 1, 2, 2, 2 };
int k = 2;
Console.WriteLine(NumberOfSubarrays(nums, k));
}
static int NumberOfSubarrays(int[] nums, int k)
{
var record = new Dictionary<int, int>();
record[0] = 1;
int numOfOdd = 0;
for (int i = 0; i < nums.Length; i++)
{
if (nums[i] % 2 != 0)
numOfOdd++;
if (record.ContainsKey(numOfOdd))
record[numOfOdd]++;
else
record[numOfOdd] = 1;
}
int res = 0;
foreach (var kv in record)
{
if (record.ContainsKey(kv.Key - k))
res += record[kv.Key - k] * kv.Value;
}
return res;
}
}
}