LeetCode 3134 找出唯一性数组的中位数

方法1：二分 + 滑动窗口 + 哈希表

class Solution {
    public int medianOfUniquenessArray(int[] nums) {
        int n = nums.length;
        // 左中位数下标 下标从 1 开始
        long median = ((long) n * (n + 1) / 2 + 1) / 2;
        int left = 1, right = n;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (check(nums, median, mid)) 
                right = mid;
            else 
                left = mid + 1;
        }
        return left;
    }

    public boolean check(int[] nums, long median, int mid) {
        // HashMap 统计不同元素个数 保证其不超过 mid
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        int left = 0; long cnt = 0; // 统计满足添加的区间个数
        for (int right = 0; right < nums.length; right++) {
            map.merge(nums[right], 1, Integer::sum);
            while (map.size() > mid) {
                if (map.merge(nums[left], -1, Integer::sum) == 0)
                    map.remove(nums[left]);
                left++; // 滑动左区间
            }
            // 前面区间已统计 此处只统计以 nums[right] 结尾的子数组
            cnt += right - left + 1;
        }
        return cnt >= median;
    }
}

class Solution:
    def medianOfUniquenessArray(self, nums: List[int]) -> int:
        def check(num: int) -> bool:
            dic = Counter(); left = 0; cnt = 0
            for right, x in enumerate(nums):
                dic[x] += 1
                while len(dic) > num:
                    dic[nums[left]] -= 1
                    if dic[nums[left]] == 0:
                        del dic[nums[left]]
                    left += 1
                cnt += right - left + 1
                if cnt >= medium: # 提前剪枝
                    return True
            return False
        
        n = len(nums)
        medium = (n * (n + 1) // 2 + 1) // 2
        low = 1; high = n
        while low < high:
            mid = (low + high) >> 1
            if check(mid):
                high = mid
            else:
                low = mid + 1
        return low