Leetcode - Contains Duplicate II

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 difference between i and j is at most k.

[分析]
思路1：使用HashSet, HashSet中存最近访问的 k 个元素，从第 k + 1个元素开始遍历检查，若 HashSet中存在同当前访问元素相同的说明 k 邻域内存在重复元素，否则添加当前元素并删除与其下标距离 为 k 的元素。特别注意若先删除再添加需要特殊处理 k == 0的情况，不然遍历过程中HashSet只增不减。
思路2：使用HashMap，key是数组元素值，value是该元素最近被访问的下标。遇到相同元素check其与上一次位置的距离是否 <= k

public class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        if (nums == null || nums.length < 2)
            return false;
        HashSet<Integer> set = new HashSet<Integer>();
        int end = Math.min(k, nums.length);
        for (int i = 0; i < end; i++) {
            if (set.contains(nums[i]))
                return true;
            set.add(nums[i]);
        }
        for (int i = k; i < nums.length; i++) {
            if (set.contains(nums[i]))
                return true;
            set.add(nums[i]);
            set.remove(nums[i - k]);
        }
        return false;
    }
    public boolean containsNearbyDuplicate2(int[] nums, int k) {
        if (nums == null || nums.length < 2)
            return false;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(nums[i])) {
                if (i - map.get(nums[i]) <= k)
                    return true;
            } 
            map.put(nums[i], i);
        }
        return false;
    }
}