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].

使用哈希表记录每一个数字的下标可以高效地获得目标数字地下标。另外unordered_map底层使用哈希表实现，map底层使用红黑树实现，所以unordered_map查找效率高于map，但是占用空间更大。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> data;
        vector<int> res;
        int len = nums.size();
        for(int i = 0; i < len; i++){
            data[nums[i]] = i;
        }
        for(int i = 0; i < len; i++){
            if(data.find(target - nums[i]) != data.end() && data[target-nums[i]] != i){
                return {i,data[target - nums[i]]};
            }
        }
        return res;
    }
};