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
