LeetCode - 兩數之和

描述：

給定一個整數數組 nums 和一個目標值 target，請你在該數組中找出和爲目標值的那 兩個 整數，並返回他們的
數組下標。
你可以假設每種輸入只會對應一個答案。但是，你不能重複利用這個數組中同樣的元素。
給定 nums = [2, 7, 11, 15], target = 9
因爲 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

def twoSumOne(nums,target):
    l = len(nums)
    for num in range(l):
        one = nums[num]
        for num1 in range(l):
            if(num == num1):
                continue
            two = nums[num1]
            if((one + two) == target):
                return [num,num1]

def twoSumTwo(nums,target):
    l = len(nums)
    for i in range(l):
        one = nums[i]
        two = target - one
        if(two not in nums):
            continue
        j = nums.index(two)
        if(j == i):
            continue
        if(j != None):
            return [i,j]

def twoSumThree(nums,target):
    l = len(nums)
    dictOne = {nums[i]:i for i in range (l)}
    dictTwo = {i:target - nums[i] for i in range(l)}
    for j in range(l):
        index = dictOne.get(dictTwo.get(j))
        if(index != None):
            return [j,index]

if __name__ == "__main__":
    parser = argparse.ArgumentParser(description="two num add")
    parser.add_argument('-l',dest='list',metavar='N',type=int,nargs='+',help='intergers for two add')
    args = parser.parse_args()
    print(args.list)
    print(twoSumTwo(args.list,10))

源碼：
命令行參數解析參考