1. Two Sum【easy】

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：

 class Solution {
 public:
     vector<int> twoSum(vector<int>& nums, int target) {
         unordered_map<int, int> u_map;
         vector<int> result;

         for (int i = ; i < nums.size(); ++i)
         {
             if (u_map.find(target - nums[i]) != u_map.end())
             {
                 result.push_back(u_map[target - nums[i]]);
                 result.push_back(i);
                 return result;
             }

             u_map.insert(make_pair(nums[i], i));
         }

         return result;
     }
 };

map搞一把

解法二：

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

比较简洁的写法

扩展见：

167. Two Sum II - Input array is sorted【easy】

170. Two Sum III - Data structure design【easy】