LeetCode第1题：Two Sum

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].

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Analyze: 

1. 注意题目里的假设：each input would have exactly one solution，因此输出答案是唯一的，不会出现[ 3, 4, 5, 6 ]这样的数组：（3+6 = 4+5 = 9），所以优先考虑的是一次遍历。
2. 在一次遍历的同时，我们可以计算出这个数需要的另一个加数，记为checkNum，例如5的checkNum就是（9-5=）4。
3. 判断当前数字是否在checkNum中，例如遍历到数字4的时候，发现有checkNum中恰好有一个4，则成功找到。

这里的难点就在于如何返回checkNum对应那个数字的下标，解决办法是：

创建一个solution_dict字典，字典的key是checkNum，value是当前数字在nums中的下标

Code:

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        solution_dict = {}
        for i in range(len(nums)):
            if (nums[i] in solution_dict):  # if already exists
                return [solution_dict[nums[i]],i]
            else:   # renew the dictionary
                # calcu what we need 
                checkNum = target - nums[i]
                solution_dict[checkNum] = i

Result:

Runtime: 40 ms, faster than 86.49% of Python3 online submissions for Two Sum.

Memory Usage: 14.5 MB, less than 5.08% of Python3 online submissions for Two Sum.