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，时间复杂度O（n^2），空间复杂度O（1）

class Solution {public:    vector
     
       twoSum(vector
      
       & nums, int target) {        vector
       
         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,其实是c++中的map（由于c++中不存在hash_table,所以没有基于hash_table实现 ）,时间复杂度可降至O(nlogn), 但是增加了空间复杂度O(n)

注意的一点是，存入map中的Key 是nums中的值，存入的val是nums中对应的索引。

 

class Solution {public:    vector
      
        twoSum(vector
       
        & nums, int target) {        vector
        
          res;        int tmp;        map
         
           hashtable;        for (int i = 0; i < nums.size(); i++){            hashtable[nums[i]] = i;       // 首先循环遍历一次将所有的值存入hashtable中        }        for (int i = 0; i < nums.size(); i++){            tmp = target -  nums[i];            if (hashtable.find(tmp)!= hashtable.end() && hashtable[tmp]!= i){  // 若是能查找到当前值相对于target的补集                res.push_back(i);                res.push_back(hashtable[tmp]);                   return res;            }        }                    }};