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].

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1. 暴力搜索，時間復雜度O（n^2），空間復雜度O（1）

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

2. 利用hashtable,其實是c++中的map（由於c++中不存在hash_table,所以沒有基於hash_table實現 ）,時間復雜度可降至O(nlogn), 但是增加了空間復雜度O(n)

註意的一點是，存入map中的Key 是nums中的值，存入的val是nums中對應的索引。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        int tmp;
        map<int, int> hashtable;
        for (int i = 0; i < nums.size(); i++){
            hashtable[nums[i]] = i;       // 首先循環遍歷一次將所有的值存入hashtable中
        }
        for (int i = 0; i < nums.size(); i++){
            tmp = target -  nums[i];
            if (hashtable.find(tmp)!= hashtable.end() && hashtable[tmp]!= i){  // 若是能查找到當前值相對於target的補集
                res.push_back(i);
                res.push_back(hashtable[tmp]);   
                return res;
            }
        }
        
        
    }
};

Leetcode -- 1. two sum