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].

給出一個integer數列和一個integer，return一個包含 在給出的數列中兩個數之和是給出的integer的index的 數列。只有唯一解。

這是一個最簡單的方法，但是runtime太高。

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for p1 in range(len(nums)):
            for
 
 p2 in range(p1+1,len(nums)):
                if((nums[p1]+nums[p2]) == target):
                    return [p1,p2]

用雜湊runtime會快很多因為只有一個loop。

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
 

        if len(nums) <= 1:
            return False
        buff_dict = {}
        for i in range(len(nums)):
            if nums[i] in buff_dict:
                return [buff_dict[nums[i]], i]
            else:
                buff_dict[target - nums[i]] = i

當list的數不在dict的時候，target減這個數的值為key，value為此數。當list的數為在dict為key的時候查詢就完成了。