LeetCode:1. Add Two Numbers

题目:
LeetCode:1. Add Two Numbers

描述：

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.

样例：

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

分析：

• 思路如下：
  1) 利用hashmap来存储数组元素以及其对应索引值，提高检索效率
  2) 通过nTarget - vecNum[i]来获取对应的检索目标元素
  3) 返回满足条件的索引值

代码：

vector<int> twoSum(vector<int>& vecNum, int nTarget) {
    vector<int> vecTemp;
    unordered_map<int, int> hashMapTemp;

    for (int i = 0; i < vecNum.size(); i++)
    {
        hashMapTemp[vecNum[i]] = i;
    }

    for (int i = 0; i < vecNum.size(); i++)
    {
        const int nPart = nTarget - vecNum[i];
        if (hashMapTemp.find(nPart) != hashMapTemp.end() && hashMapTemp[nPart] > i)
        {
            vecTemp.push_back(i);
            vecTemp.push_back(hashMapTemp[nPart]);
            break;
        }
    }

    return vecTemp;
    }

备注：
LC上大神的 0ms 的代码：

vector<int> twoSum(vector<int> &numbers, int target)
{
    //Key is the number and value is its index in the vector.
	unordered_map<int, int> hash;
	vector<int> result;
	for (int 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()) {
                    //+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;
}