217/219. Contains Duplicate /Contains Duplicate II

原文题目：

217. Contains Duplicate

219. Contains Duplicate II

读题：

217只要找出是否有重复值，

219找出重复值，且要判断两者索引之差是否小于k

//217. Contains Duplicate 46ms
class Solution
{
public:
	bool containsDuplicate(vector<int>& nums)
	{
		set <int> s;
		//vector <int>::iterator it;
		int i = 0;
		if(nums.empty())
		{
			return false;
		}
		s.insert(nums[0]);
		for(i = 1;i < nums.size();++i)
		{
			if(s.count(nums[i]))
			{
				return true;
			}
			s.insert(nums[i]);
		}
		return false;
	}
};

//217. Contains Duplicate 29ms
class Solution {
public:
	bool containsDuplicate(vector<int>& nums) {
		std::sort(nums.begin(), nums.end());
		int i = 0, j = nums.size() - 1;
		while (i < j) {
			if (nums[i] == nums[i+1])
				return true;
			++i;
		}
		return false;
	}
};

//219. Contains Duplicate II 25ms
class Solution {
public:
	bool containsNearbyDuplicate(vector<int>& nums, int k) {
		unordered_map <int, int> m; //如果用的是map，那么时间复杂度为35ms
		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;
	}
};