Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

Example 1:

Input: nums = [1,2,3,1], k = 3

Output: true

Example 2:

Input: nums = [1,0,1,1], k = 1

Output: true

Example 3:

Input: nums = [1,2,3,1,2,3], k = 2

Output: false

217. Contains Duplicate 的拓展，那道题只需判断数组中是否有重复值，而这题限制了数组中只许有一组重复的数字，而且坐标差最多k。

解法：哈希表Hash table

Java:

class Solution {

    public boolean containsNearbyDuplicate(int[] nums, int k) {

        HashMap<Integer, Integer> index = new HashMap<>();

        for (int i = 0; i < nums.length; i++) {

            if (i - index.getOrDefault(nums[i], -k - 1) <= k) return true;

            index.put(nums[i], i);

        }

        return false;

    }

}

Python:

class Solution:

    # @param {integer[]} nums

    # @param {integer} k

    # @return {boolean}

    def containsNearbyDuplicate(self, nums, k):

        lookup = {}

        for i, num in enumerate(nums):

            if num not in lookup:

                lookup[num] = i

            else:

                # It the value occurs before, check the difference.

                if i - lookup[num] <= k:

                    return True

                # Update the index of the value.

                lookup[num] = i

        return False

C++:

class Solution {

public:

    bool containsNearbyDuplicate(vector<int>& nums, int k) {

        unordered_map<int, int> m;

        for (int i = 0; i < nums.size(); ++i) {

            if (m.find(nums[i]) != m.end() && i - m[nums[i]] <= k) return true;

            else m[nums[i]] = i;

        }

        return false;

    }

};

类似题目：

[LeetCode] 217. Contains Duplicate 包含重复元素

[LeetCode] 220. Contains Duplicate III 包含重复元素 III

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巴特西

[LeetCode] 219. Contains Duplicate II 包含重复元素 II

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