1. Two Sum

1. Two Sum

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

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

人家媳妇：
time \(O(n)\), space \(O(n)\)
方法中巧妙使用map，一次遍历完成任务．
遍历数组时，若未在map中找到想要的元素，则把数组中当前元素投入map．(注key=数组元素值，val=该元素indx)

vector<int> twoSum(vector<int> &numbers, int target) {
	//Key is the number and value is its index in the vector.
	map<int, int> hash; //原来人家用的是unordered_map
	vector<int> result;
	for (int unsigned i = 0; i < numbers.size(); i++) {
		int numberToFind = target - numbers[i];

		//if numberToFind is found in map, return them
		if (hash.find(numberToFind) != hash.end()) {
      //map 中若找到
			//+1 because indices are NOT zero based
			result.push_back(hash[numberToFind] + 1);
			result.push_back(i + 1);
			return result;
		}

		//number was not found. Put it in the map.
		hash[numbers[i]] = i;
	}
	return result;
}

自家媳妇 😃
time \(O(n^2)\), space \(O(1)\)
Brute Force

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