1. Two Sum

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。