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 absolute difference between i and j is at most k.

思想：创建一个集合，在这个集合中维护k个元素，这k个元素是相对于当前遍历的元素的前k个元素，当遍历一个新的元素时，检查前k个元素中是否有相同的元素出现。

#include<iostream>#include<unordered_set>#include<vector>using namespace std;class Solution {public:    bool containsNearbyDuplicate(vector<int>& nums, int k) {        unordered_set<int> preKValueSet;        if(k <= 0)            return false;        if(k >= nums.size())            k = nums.size() - 1;        for(int i = 0; i < nums.size(); i++)        {            if(i > k)                preKValueSet.erase(nums[i - k - 1]);            if(preKValueSet.find(nums[i]) != preKValueSet.end())                return true;            preKValueSet.insert(nums[i]);        }        return false;    }};

还有一种使用哈希表的方式如下

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;    }};