2019.4.14 LeetCode-Two Sum

what is “two sum” problem?
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
给定一个整数数组，返回两个数字的索引，使它们相加到特定目标。

solution1: Brute Force蛮力法

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) 
{
	vector<int> res;
   for(int i=0;i<(nums.size()-1);i++)
   {
	   for (int j = i + 1; j < nums.size(); j++)
	   {
		   if (nums[i] + nums[j] == target)
		   {
			   res.push_back(i);
			   res.push_back(j);
			   break;
		   }
	   }
   }  
   return res;
}
};

Time complexity : O(n^2) ，Space complexity : O(1)

solution2：Hash Table哈希表
What is the best way to maintain a mapping of each element in the array to its index? A hash table.哈希表是获取数组中某个指定元素的索引的最好方法。
We reduce the look up time from O(n)O(n) to O(1)O(1) by trading space for speed. 通过增加空间消耗来较少时间消耗。
Beware that the complement must not be nums[i]nums[i] itself!注意获取的另一个元素不能和第一个元素是同一个数。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) 
{
	map<int, int> m;
	vector<int>res;
	for (int i = 0; i < nums.size(); i++)
	{
		pair<int, int> p(nums[i], i);
		m.insert(p);
	}
	for (int i = 0; i < nums.size(); i++)
	{
		int complement = target - nums[i];
		if (m.count(complement) &&((m.find(complement)->second)!=i))
		{
			res.push_back(i);
			res.push_back((m.find(complement)->second));
            break;
		}
	}
    return res;
}
};

注意：map.count()和map.find()的区别：map.count()是判断hash表中存不存在某个值，如果存在返回1，反之为0。map.find()是返回指向这个值的指针，可以通过指针访问对应的键值。

Time complexity : O(n) ，Space complexity : O(n)

solution3：优化solution2

通过改变给map赋初值的顺序，可以有效回避返回两个元素相同的情况。

class Solution {
public:
   vector<int> twoSum(vector<int>& nums, int target) {
		vector<int>res;
		map<int, int> m;
		for (int i = 0; i < nums.size(); i++)
		{
			int complement = target - nums[i];
			if (m.count(complement))
			{
				res.push_back(m.find(complement)->second);
				res.push_back(i);
				return res;
			}
			pair<int, int> p(nums[i], i);
			m.insert(p);
		}
        return res;
	}
}; 

Time complexity : O(n) ，Space complexity : O(n)