[Leetcode1_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].

求整型数组中，和为某个数（target）的两个数的下标

解法一：暴力循环, 复杂度为O(n^2)

解法二：建立字典，利用字典的内置的查找函数降低复杂度

#PYTHON 代码:
class Solution():
def TwoSum(nums, target):
d = {}
for i in range(len(nums)):
if d[nums[i]] not in d:
d[target-nums[i]] = i
else:
return d[nums[i]], i

//c++解法
#include <unordered_map>
class Solution{
vector<int> TwoSum(vector<int> &nums, int target){
std::unordered_map<int, int> d;
vector<int> res;
for(int i = 0; i < nums.size(); ++i){
if(d.find(nums[i] == d.end()){
d[target - nums[i]] = i;
}else{
res.push_back(d[nums[i]]);
res.push_back(i);
return res;
}
}
}
}