给定一个整数数组和一个整数 k，判断数组中是否存在两个不同的索引 i 和 j，使得 nums [i] = nums [j]，并且 i 和 j 的差的 绝对值 至多为 k。 示例 1:输入: nums = [1,2,3,1], k = 3输出: true示例 2:输入: nums = [1,0,1,1], k = 1输出: true示例 3:输入: nums = [1,2,3,1,2,3], k = 2输出: false

/** * @param {number[]} nums * @param {number} k * @return {boolean} */var containsNearbyDuplicate = function (nums, k) {  const visited = {};  for (let i = 0; i < nums.length; i++) {    const num = nums[i];    if (visited[num] !== undefined && i - visited[num] <= k) {      return true;    }    visited[num] = i;  }  return false;};

class Solution:    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:        d = {}        for index, num in enumerate(nums):            if num in d and index - d[num] <= k:                return True            d[num] = index        return False

class Solution {public:    bool containsNearbyDuplicate(vector<int>& nums, int k) {        auto m = unordered_map<int, int>();        for (int i = 0; i < nums.size(); ++i) {            auto iter = m.find(nums[i]);            if (iter != m.end()) {                if (i - m[nums[i]] <= k) {                    return true;                }            }            m[nums[i]] = i;        }        return false;    }};

class Solution {    public boolean containsNearbyDuplicate(int[] nums, int k) {        Map<Integer, Integer> map = new HashMap<>();        for(int i=0;i<nums.length;i++)        {            if(map.get(nums[i]) != null && (i-map.get(nums[i])) <= k)            {                return true;            }            map.put(nums[i], i);        }        return false;    }}