LeetCode 1. Two Sum

Problem: 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].

该题最容易想到的暴力解法就是两层循环,逐对相加与target比较。其代价过大，代码为：

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

时间复杂度为O(n)的做法是只遍历一次vector，然后用hash表的find快速查找存储的数字。其代码如下：

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

已遇到多题巧用hash table(unordered_map)简化算法的方法，注意hash table的使用！