Q1 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].

思路：

将列表的索引和值保存到map中，判断差值是否也存在的列表中，可以达到O(n)复杂度。

Python实现：

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        map = {}
        for i, val in enumerate(nums):  # 遍历列表的下标和索引，也可以使用 for i in range(len(nums))
            gap = target - val
            if gap in map:  # 如果键存在于map中
                return [map[gap], i]
            else:
                map[val] = i  # 按照{值:索引}的形式保存到map中
        return None

a = [2,3,5,7,11]
b = Solution()
print(b.twoSum(a[:], 16))  # [2, 4]