1. Two Sum

Difficulty: 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.

Example:

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

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

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

假设一定存在数组其中的2个数相加等于目标值，那么只需要循环数组，2个for循环嵌套一个个的相加得到和数等于目标值返回即可，于是就是第一种方法。复杂度O(n2)，空间复杂度O(1).

My submission 1: Accepted  Runtime: 648ms

 public int[] TwoSum(int[] nums, int target) {
         int[] result=new int[2];
            for (int i = 0; i<nums.Length; i++)
            {                
                    for (int j = i + 1; j < nums.Length; j++)
                    {
                        if (nums[i] + nums[j] == target)
                        {
                            result[0] = i;
                            result[1] = j;
                            return result;
                        }
                    }               
            }
            return result;  
    }

考虑到第一个方法的O(n2)的复杂度，我们需要改良，循环存数进字典，倘若里面有值等于目标值-现在的值那么即匹配，返回2个坐标值。时间复杂度O(n)，空间复杂度O(n)。

My submission 2: Accepted Runtime: 476ms

public int[] TwoSum(int[] nums, int target) {
          Dictionary<int, int> dic=new Dictionary<int, int>();
            for (int i = 0; i < nums.Length; i++)
            {
                if(dic.ContainsKey(target-nums[i]))
                    return new int[] {dic[target-nums[i]],i};
                else if(!dic.ContainsKey(nums[i]))
                {
                    dic.Add(nums[i],i);
                }
            }
            return new int[] {0,0};
    }