思路：
- 简单点的，暴力就完事了，但是会超时
- 利用滑动窗口思想，利用List模拟一个窗口，在窗口中利用map存储数字出现的次数，每次窗口移动遍历map判断是否有数出现的次数达到阈值t
//暴力法 只能通过82% public static void main(String[] args) { Scanner in = new Scanner(System.in); while (in.hasNext()) { Map map = new TreeMap<>(); int res = 0; int n = in.nextInt(); int k = in.nextInt(); int t = in.nextInt(); int[] arr = new int[n]; for (int i = 0; i < n; i++) { arr[i] = in.nextInt(); } for (int i = 0, j = i + k - 1; j < n; i++, j++) { for (int p = i; p <= j; p++) { if (map.containsKey(arr[p])) { map.put(arr[p], map.get(arr[p]) + 1); } else { map.put(arr[p], 1); } } if (isT(map, t)) { res++; } map.clear(); } System.out.println(res); } } private static boolean isT(Map map,int t) { for (Integer key : map.keySet()) { int value = map.get(key); if (value >= t) { return true; } } return false; } //滑动窗口法public class Main { static class Node { int index; int value; Node(int index, int value) { this.index = index; this.value = value; } } public static void main(String[] args) { Scanner in = new Scanner(System.in); while (in.hasNext()) { int n = in.nextInt(); int k = in.nextInt(); int t = in.nextInt(); int[] arr = new int[n]; for (int i = 0; i < arr.length; i++) { arr[i] = in.nextInt(); } //用一个List来模拟窗口 LinkedList list = new LinkedList<>(); //value：times利用map来统计每个value出现的次数 HashMap map = new HashMap<>(); int end = 0; int res = 0;//统计次数 while (end < arr.length) { if (list.size() < k) { list.add(new Node(end, arr[end])); addNode(map,arr,end); end++; } if (list.size() == k) { //统计出现的次数有没有大于t int temptimes = 0; for (Map.Entry entry : map.entrySet()) { temptimes = Math.max(temptimes, entry.getValue()); } if (temptimes >= t) { res++; } //弹出list最前面的值 Node node = list.poll(); if (map.get(node.value) == 1) { map.remove(node.value); }else { //出现次数-1 map.put(node.value, map.get(node.value) - 1); } } } System.out.println(res); } } private static void addNode(HashMap map,int arr[],int end) { if (!map.containsKey(arr[end])) { map.put(arr[end], 1); }else { map.put(arr[end], map.get(arr[end])+1); } }}