【Leetcode】1. Two Sum

传送门

【题意】：从给定的一组数中找到两个数的和为target.返回两个数的序号，假设每次输入只有一组解

代码1：

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

 代码2：

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;

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