992.K个不同整数的子数组

思路：

class Solution {public int subarraysWithKDistinct(int[] nums, int k) {return atMostKDistinct(nums,k) - atMostKDistinct(nums,k-1);}//最多包含K个不同整数的子区间个数private int atMostKDistinct(int[] a,int k){int len = a.length;int[] freq = new int[len + 1];int left = 0,right = 0;int count = 0,res = 0;//[left,right)里不同整数的个数while(right < len){if(freq[a[right]] == 0){count++;}freq[a[right]]++;right++;while(count > k){freq[a[left]]--;if(freq[a[left]] == 0){count--;}left++;}res += right - left;}return res;}
}