1.Two Sum(c++)(附6ms O(n) accepted 思路和代码)

问题描述：

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

思路1：最简单的思路就是暴力求解，即循环遍历所有的可能情形，复杂度为o(n^2). 耗时未知，leetcode: Time Limit Exceeded

 class Solution {
 public:
     vector<int> twoSum(vector<int>& nums, int target) {
         vector<int> result;
         int n = nums.capacity();
         for(int i=;i<n;++i)
         {
             int i1 = nums.at(i);
             for(int j=i+;j<n;++j)
             {
                 if(i1+nums.at(j)==target)
                 {
                     result.push_back(i);
                     result.push_back(j);
                     return result;
                 }
             }
         }
         return result;
     }
 };

思路2：使用hash map，查询等各种操作都是常数级别的时间复杂度（最好不要使用map，map实现大都是使用红黑树，查询的时间复杂度是O(logn)),复杂度为O(n),leetcode: AC,16ms

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

思路3（参考论坛中4ms的c代码）：别出机杼，逆向思考，第一步找出数组中与目标值的差值（这一步可以筛选部分明显不可能得到目标值的数），第二步则是遍历数组找到哪个数是差值。复杂度：O(n). leetcode:AC,6ms,超过99.57%的结果！

 class Solution {
 public:
     vector<int> twoSum(vector<int>& nums, int target) {
         vector<int> result;
         int n = nums.size();
         int max=, min=;
         for (int i = ; i < n; i++)
         {
             if (nums[i] > max)
                 max = nums[i];
             if (nums[i] < min)
                 min = nums[i];
         }

         int* repeat_tag = new int[max - min + ]();
         int diff;
         for (int i = ; i < n; i++)
         {
             diff = target - nums[i];
             if (diff <= max && diff >= min)
             {
                 repeat_tag[diff - min]= i;
             }
         }
         for (int i = ; i < n; i++)
         {
             if (repeat_tag[nums[i] - min] != )
             {
                 if (i != repeat_tag[nums[i] - min])
                 {
                     result.push_back(i);
                     result.push_back(repeat_tag[nums[i] - min]);
                     return result;
                 }
             }
         }
         return result;
     }
 };