leetcode 1. 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].

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        num_maps = {}
        for index, value in enumerate(nums):
            complement = target - value
            complement_index = num_maps.get(complement)
            if complement_index is not None:
                return [complement_index, index]
            num_maps.update({value: index})
        return []

思路：
其实很简单，题目已经假定肯定有匹配数字，所以不用考虑其它情况。

直接搞个dict，每次遍历一个数就把它和它的索引存在里面，然后每次都检查补数在不在dict里面：

如果在，返回补数索引和本次遍历的数字索引。

Complexity Analysis:

Time complexity : O(n). We traverse the list containing n elements only once. Each look up in the table costs only O(1) time.

Space complexity : O(n). The extra space required depends on the number of items stored in the hash table, which stores at most n elements.