leetcode_No.1 Two Sum

原题：　　

　　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.

翻译：

　　给出一个数字列表和一个目标值（target），假设列表中有且仅有两个数相加等于目标值，我们要做的就是找到这两个数，并返回他们的索引值。

举例：

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

 

解答：　

# 方法1：双层循环，面试不会通过，时间复杂度太高
def twoSum_way1(self,nums,target):
    """
    :param nums: 整数列表
    :param target:目标值
    :return:result 索引值
    """
    result = []
    for i in range(len(nums)):
        for j in range(i+1,len(nums)):
            if nums[i] + nums[j] == target:
                result.append(i)
                result.append(j)
                return result

#方法2：通过判断target与某一个元素的差值是否也在列表之中,类似方法1,同样是面试官不期望的回答
def twoSum_way2(self,nums,target):
    """
    :param nums: 整数列表
    :param target:目标值
    :return:result 索引值
    """
    result = []
    for i in range(len(nums)):
        first_num = nums[i]
        second_num = target - first_num
        if second_num in nums:
            j = nums.index(second_num)
            if i != j:
                result.append(i)
                result.append(j)
                return result

　　

 

#适合面试的方法：
#方法3：通过创建字典，将nums里的值和序号对应起来，
# 并创建另一个字典存储目标值（Target）-nums的值，
# 通过判断该值是否在nums内进行判断并返回其对应索引值
class Solution:
    def twoSum_way3(self, nums, target):
        """
        :param nums: 整数列表
        :param target:目标值
        :return:result 索引值
        """
        #创建第一个字典：用于存储整数列表nums的元素值和对应索引
        num_dict = {nums[i]: i for i in range(len(nums))}
        #创建第二个字典：存储target-列表中的元素的值
        num_dict2 = {i: target - nums[i] for i in range(len(nums))}
        #判断num_dict2的值是否是输入列表中的元素，如果是返回索引值，不是则往下进行
        result = []
        for i in range(len(nums)):
            j = num_dict.get(num_dict2.get(i))
            if (j is not None) and (j != i):
                result = [i,j]
                break
        return result

#方法4:改进方法3，让代码简洁一些
class Solution:
    def twoSum_way4(self, nums, target):
        """
        :param nums: 整数列表
        :param target:目标值
        :return:result 索引值
        """
        num_dict = {nums[i]: i for i in range(len(nums))}
        for i in range(len(nums) -1):
            difference = target - nums[i]
            if difference in num_dict and i != num_dict[difference]:
                return [i,num_dict[difference]]
        return None

 

终极写法：

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dict = {}
        for index,num in enumerate(nums):
            another_num = target - num
            dict[num] = index
            if another_num in dict:
                return [dict[another_num], index]
        return None