题解 | 相差不超过k的最多数

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n, k;
    cin >> n >> k;
    
    vector<int> nums(n);
    for(int i = 0; i < n; i++) {
        cin >> nums[i];
    }
    
    // 排序
    sort(nums.begin(), nums.end());
    
    // 滑动窗口
    int maxCount = 1;
    int left = 0;
    
    for(int right = 1; right < n; right++) {
        // 如果当前窗口的最大差值超过k，移动左指针
        while(left < right && nums[right] - nums[left] > k) {
            left++;
        }
        maxCount = max(maxCount, right - left + 1);
    }
    
    cout << maxCount << endl;
    return 0;
}

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        
        int[] nums = new int[n];
        for(int i = 0; i < n; i++) {
            nums[i] = sc.nextInt();
        }
        
        // 排序
        Arrays.sort(nums);
        
        // 滑动窗口
        int maxCount = 1;
        int left = 0;
        
        for(int right = 1; right < n; right++) {
            // 如果当前窗口的最大差值超过k，移动左指针
            while(left < right && nums[right] - nums[left] > k) {
                left++;
            }
            maxCount = Math.max(maxCount, right - left + 1);
        }
        
        System.out.println(maxCount);
    }
}

n, k = map(int, input().split())
nums = list(map(int, input().split()))

# 排序
nums.sort()

# 滑动窗口
max_count = 1
left = 0

for right in range(1, n):
    # 如果当前窗口的最大差值超过k，移动左指针
    while left < right and nums[right] - nums[left] > k:
        left += 1
    max_count = max(max_count, right - left + 1)

print(max_count)