【Leetcode】Contains Duplicate II

题目链接：https://leetcode.com/problems/contains-duplicate-ii/

题目：

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.

思路：

1、用HashMap保存元素的下标，如果出现重复元素，则判断是否满足 i-j<=k，不满足则更新HashMap保存的下标，因为后面的元素可能还有重复，要跟最近的重复元素比较距离。

2、用Set，具体看代码。。。时间久了我也忘了具体怎么做的。。。。

算法1：

public boolean containsNearbyDuplicate(int[] nums, int k) {
HashMap<Integer, Integer> set = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; i++) {
if (!set.containsKey(nums[i])) {
set.put(nums[i], i);// 记录起始位置
} else {
if (i - set.get(nums[i]) <= k) {
return true;
} else {
set.put(nums[i], i);
}
}
}
return false;
}

算法2：

public boolean containsNearbyDuplicate(int[] nums, int k) {
Set<Integer> set = new HashSet<Integer>();
int start = 0, end = 0;
for (int i = 0; i < nums.length; i++) {
if (!set.contains(nums[i])) {
set.add(nums[i]);
end++;
} else
return true;
if (end - start > k) {
set.remove(nums[start]);
start++;
}
}
return false;
}