1. Two Sum

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题目：

https://leetcode.com/problems/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.

1. 最直接实现，冒泡法，两两相加。（算法复杂度o(n*n）不能通过)

c++实现方法

vector<int> twoSum(vector<int>& vec, int target) {         vector<int>::iterator it;        for(it=vec.begin();it!=vec.end();it++){            vector<int>::iterator it2;            for(it2=it+1;it2!=vec.end();it2++){                if(*it + *it2 == target){                    vector<int> re;                    re.push_back(it-vec.begin());                    re.push_back(it2-vec.begin());                    return re;                }            }        }        return vector<int>();    }

2. 使用map实现，算法复杂度o(n*logn)

C++实现

vector<int> twoSum(vector<int>& vec, int target) {        map<int ,int>mapTarget;         for(int i = 0; i < vec.size(); i++){             if(mapTarget.find(vec[i]) != mapTarget.end()){                vector<int> re;                 re.push_back(mapTarget.find(vec[i])->second);                 re.push_back(i);                return re;            }            mapTarget.insert(pair<int,int>(target - vec[i], i));        }         return vector<int>();    }

3. 使用 hash_map来存值，算法复杂度 o(n)

python实现

def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        ret_map = {}        for i, num in enumerate(nums):            index = ret_map.get(target - num, None)            if index is None:                ret_map[num] = i            else:                return [ret_map[target - num], i]

对代码做些优化

def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        ret_map = {}        for i, num in enumerate(nums):             if target - num in ret_map:                return [ret_map[target - num], i]            ret_map[num] = i

4. 先排序，再前后杂遍历，时间复杂度o(n*logn + n)

算法复杂度偏高，题目要求把初始位置记录起来，比较麻烦，就没实现。

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