例题——2488. 统计中位数为 K 的子数组⭐

https://leetcode.cn/problems/count-subarrays-with-median-k/description/

提示：

n == nums.length
1 <= n <= 10^5
1 <= nums[i], k <= n
nums 中的整数互不相同

【套路】子数组统计问题常用技巧：等价转换

https://leetcode.cn/problems/count-subarrays-with-median-k/solutions/1993439/deng-jie-zhuan-huan-pythonjavacgo-by-end-5w11/
注意，每个数字各不相同，整个数组中只有一个 k。

class Solution {public int countSubarrays(int[] nums, int k) {int pos = 0, n = nums.length;// 求出 k 的 poswhile (nums[pos] != k) ++pos;Map<Integer, Integer> cnt = new HashMap<>();cnt.put(0, 1);for (int i = pos - 1, x = 0; i >= 0; --i) {x += nums[i] < k? 1: -1;cnt.merge(x, 1, Integer::sum);}int ans = cnt.get(0) + cnt.getOrDefault(-1, 0);for (int i = pos + 1, x = 0; i < n; ++i) {x += nums[i] > k? 1: -1;// x 和 x - 1 对应 奇数长度和偶数长度ans += cnt.getOrDefault(x, 0) + cnt.getOrDefault(x - 1, 0);}return ans;}
}

本质上是「中心前后缀 + 哈希表计数」

核心在于「个数」，能得到一个和个数有关的数学式子，就可以转换成 1（或者 -1），毕竟在讨论个数的时候，每个元素的贡献就是 1 了。

相似题目列表

面试题 17.05. 字母与数字

https://leetcode.cn/problems/find-longest-subarray-lcci/description/

前缀和设置为 字母与数字数量之差，相同前缀和的一对可以组成一个子字符串。

class Solution {public String[] findLongestSubarray(String[] array) {int st = 0, ml = 0;int n = array.length, d = 0;Map<Integer, Integer> m = new HashMap<>();m.put(0, 0);for (int i = 0; i < n; ++i) {char ch = array[i].charAt(0);if (ch >= '0' && ch <= '9') d++;else d--;if (m.containsKey(d)) {int j = m.get(d);if (i - j + 1 > ml) {st = j;ml = i - j + 1;}} else m.put(d, i + 1);}String[] ans = new String[ml];System.arraycopy(array, st, ans, 0, ml);return ans;}
}

525. 连续数组

https://leetcode.cn/problems/contiguous-array/description/

跟上面那道题几乎一模一样。

class Solution {public int findMaxLength(int[] nums) {int d = 0, ans = 0;Map<Integer, Integer> m = new HashMap<>();m.put(0, 0);for (int i = 0; i < nums.length; ++i) {if (nums[i] == 1) d++;else d--;if (m.containsKey(d)) {int id = m.get(d);if (i - id + 1 > ans) ans = i - id + 1;} else m.put(d, i + 1);}return ans;}
}

1124. 表现良好的最长时间段

https://leetcode.cn/problems/longest-well-performing-interval/description/

提示：

1 <= hours.length <= 10^4
0 <= hours[i] <= 16

解法1

class Solution {public int longestWPI(int[] hours) {int n = hours.length, d = 0, ans = 0;Map<Integer, Integer> m = new HashMap<>();m.put(0, 0);for (int i = 0; i < n; ++i) {if (hours[i] > 8) d++;else d--;if (d > 0) ans = Math.max(ans, i + 1);else if (m.containsKey(d - 1)) {ans = Math.max(ans, i - m.get(d - 1) + 1);}if (!m.containsKey(d)) m.put(d, i + 1);}return ans;}
}

解法2——利用单调栈

https://leetcode.cn/problems/longest-well-performing-interval/solutions/2110211/liang-chong-zuo-fa-liang-zhang-tu-miao-d-hysl/

class Solution {public int longestWPI(int[] hours) {int n = hours.length;int[] s = new int[n + 1];ArrayDeque<Integer> stk = new ArrayDeque<>();stk.push(0);// 单调递减的单调栈for (int i = 1; i <= n; ++i) {s[i] = s[i - 1] + (hours[i - 1] > 8? 1: -1);if (s[i] < s[stk.peek()]) stk.push(i);}int ans = 0;for (int i = n; i > 0; --i) {while (!stk.isEmpty() && s[i] > s[stk.peek()]) {ans = Math.max(ans, i - stk.pop());}}return ans;}
}