LeetCode | Contains Duplicate II

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

//要求判断数组是否有重复，且重复的两个元素的index的距离小于等于k//直接的思路是用map<nums[i], index>来记录数组值及对应的indexpublic class Solution {    public boolean containsNearbyDuplicate(int[] nums, int k) {                boolean result = false;        Map<Integer, Integer> myMap = new HashMap<Integer, Integer>(nums.length);                for(int i=0; i<nums.length; i++){            if(myMap.containsKey(nums[i]) && i-myMap.get(nums[i]) <= k){                result = true;                break;            }                        myMap.put(nums[i], i);  //无论已遍历的集合是否containsKey，当不contains时为插入<nums[i],index>        }                           //当contains时，为更新已有<nums[i],index>映射中的index值                                    //即，当后put的key在map中已经存在时，就覆盖掉之前的<k,v>对        return result;    }}

//用长度k的窗口来表述间距k的限定条件，而不是用map来记录indexpublic class Solution {    public boolean containsNearbyDuplicate(int[] nums, int k) {                boolean result = false;        Set<Integer> mySet = new HashSet<Integer>(nums.length);                for(int i=0; i<nums.length; i++){            if(mySet.contains(nums[i])){                result = true;                break;            }            mySet.add(nums[i]);            if(i>=k){                     //始终让set的大小为k，超了就将前面的元素删除                mySet.remove(nums[i-k]);            }        }                return result;    }}