1. Two Sum【easy】

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].

解法一：

1 class Solution { 2 public: 3     vector
     
       twoSum(vector
      
       & nums, int target) { 4         unordered_map
       
         u_map; 5         vector
        
          result; 6          7         for (int i = 0; i < nums.size(); ++i) 8         { 9             if (u_map.find(target - nums[i]) != u_map.end())10             {11                 result.push_back(u_map[target - nums[i]]);12                 result.push_back(i);13                 return result;14             }15             16             u_map.insert(make_pair(nums[i], i));17         }18                                  19         return result;20     }21 };
        
       
      
     

map搞一把

 

解法二：

1 class Solution { 2 public: 3     vector
     
       twoSum(vector
      
       & nums, int target) { 4         unordered_map
       
         map; 5         int n = (int)nums.size(); 6         for (int i = 0; i < n; i++) { 7             auto p = map.find(target - nums[i]); 8             if (p != map.end()) { 9                 return {p->second, i};10             }11             map[nums[i]] = i;12         }13     }14 };
       
      
     

比较简洁的写法

 

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