LC1. Two Sum

Description

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].

Solution

• Brute Force：

  遍历nums，对每个元素遍历它之后的所有元素，时间复杂度, 空间复杂度

• Hash:

  因为每个元素都是要找its complement, 所以complement可以作为一个index，对应的位置存储这个元素的位置。使用hash达到的时间复杂度，的空间复杂度。

  class Solution:
      def twoSum(self, nums: List[int], target: int) -> List[int]:
          value = {}
          for i in range(len(nums)):
              if nums[i] in value.keys():
                  return [value[nums[i]], i]
              else:
                  value[target-nums[i]] = i
  

Tips

• 实测发现使用for i in range(len(nums)) 比for i, x in enumerate(nums) 更快诶