1. Two Sum

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.

Example:

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

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

解法1：暴力法。时间复杂度为O(n^2)，空间复杂度为O(1)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        if(nums.size()<2)return res;
        for(size_t i=0; i<nums.size(); ++i)
        {
            for(size_t j=i+1; j<nums.size(); ++j)
            {
                if(nums[i]+nums[j]==target)
                {
                    res.push_back(i);
                    res.push_back(j);
                    return res;
                }
            }
        }
        return res;
    }
};

　　解法2：用一个hash表（c++中可使用unordered_map）记录下出现过的数字及其对应的位置，以空间换时间，可将时间复杂度降低到O(n)，空间复杂度为O(n)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        std::unordered_map<int, int> hashMap;
        for(size_t idx=0; idx<nums.size(); ++idx)
        {
            auto it=hashMap.find(target-nums[idx]);
            if(it!=hashMap.end())
            {
                res.push_back(it->second);
                res.push_back(idx);
                return res;
            }else{
                hashMap.insert(std::make_pair(nums[idx], idx));
            }
        }
        return res;
    }
};