# LeetCode219. Contains Duplicate II（滑动指针）

## LeetCode219. Contains Duplicate II（滑动指针）

• Easy
• Accepted：181,620
• Submissions：525,006

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

### 链接

https://leetcode.com/problems/contains-duplicate-ii/

### 题解

• 暴力解法，两层循环，时间复杂度$O(n^2)$，准确一点，应该是$O((n-k)*k)$
• 将其转化成在一个区间[l, l+k]中寻找两个相等的数，那么可以使用滑动窗口来做。因为[l, l+k]区间中总共有k+1个元素，所以我们可以保持滑动窗口的大小不超过k，最大限度将第k+1个元素和窗口中k个元素比较是否相等，即使用unordered_map来查询是否存在。

### 代码

• Runtime: 32 ms, faster than 95.45% of C++ online submissions for Contains Duplicate II.
• Memory Usage: 14.3 MB, less than 80.00% of C++ online submissions for Contains Duplicate II.
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
int n = nums.size();
unordered_set <int> uset;
for (int i = 0; i < n; ++i)
{
if (uset.find(nums[i]) != uset.end())
return true;
uset.insert(nums[i]);
if (uset.size() == k + 1) // 保持滑动窗口中始终保持k个元素
uset.erase(nums[i - k]);
}
return false;
}
};

The end.
2019年3月2日 星期六