漫长刷题路：twoSum系列

一、TwoSum返回下标

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].（返回数组下标）

思路：用哈希表降低时间复杂度

class Solution
{
public:
	vector<int> twoSum(vector<int>& nums, int target)
	{
		unordered_map<int, int> m;//哈希表
		vector<int> result;
		for (int i = 0; i < nums.size(); i++)
		{
			int tmp = target - nums[i];
			if (m.find(tmp) != m.end())
			{
				result.push_back(i);
				result.push_back(m[tmp]);
			}
			else
			{
				m[nums[i]] = i;
			}
		}
		return result;
	}
};
int main()
{
	vector<int> v = {2,7,11,15,17};
	Solution a;
	vector<int> res = a.twoSum(v, 17);//返回的是vector的下标
	for (int i : res)
	{
		cout << i << endl;
	}
}

 

二、twoSum返回第几个数字

Input: numbers = [2,7,11,15], target = 9

Output: [1,2]

Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

思路：前后两个指针一起移动降低时间复杂度

class Solution
{
public:
	vector<int> twoSum(vector<int>& nums,int target)
	{
		vector<int> result;
		int p1 = 0;
		int p2 = nums.size()-1;
		while (p1 < p2)
		{
			if (nums[p1] + nums[p2] > target)
			{
				p2--;
			}
			if (nums[p1] + nums[p2] < target)
			{
				p1++;
			}
			result.push_back(p1 + 1);
			result.push_back(p2 + 1);
			return result;
		}
		return result;		
	}
};

三、BST树版本twoSum

思路：将BST树转换成vector容器

typedef struct TreeNode//BST树
{
	int val;
	TreeNode* left;
	TreeNode* right;
	TreeNode(int x) :val(x), left(NULL), right(NULL)
	{}
}TreeNode;
class Solution
{
public:
	bool findTarget(TreeNode* root, int target)
	{
		vector<int> tmp;//用于存放结点的值
		inorder(root, tmp);
		int p1 = 0;
		int p2 = tmp.size() - 1;
		while (p1 < p2)
		{
			int sum = tmp[p1] + tmp[p2];
			if (sum < target)
			{
				p1++;
			}
			if (sum > target)
			{
				p2--;
			}
			return true;
		}
		return false;
	}
private:
	void inorder(TreeNode* root, vector<int>& a)//把BST树转换成vector是从小到大的顺序
	{
		if (root == nullptr)
			return;
		inorder(root->left, a);
		a.push_back(root->val);
		inorder(root->right,a);
	}
};