Leetcode：1.Two Sum

摘要： 這是崔斯特的第八十九篇原創文章 1. Two Sum 難度: Easy 刷題內容 原題連線： https://leetcode.com/problems/two-sum 內容描述： Given an array of integers, return ...

這是崔斯特的第八十九篇原創文章

1. Two Sum

難度: Easy

刷題內容

原題連線： https://leetcode.com/problems/two-sum

內容描述：

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].

解題方案

思路 1

- 時間複雜度: O(N^2)- 空間複雜度: O(1) *

暴力解法，兩輪遍歷

beats 12.80%

class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if nums[i] + nums[j] == target:
return [i, j]

Runtime: 5248 ms, faster than 12.80% of Python3 online submissions forTwo Sum.
Memory Usage: 13.5 MB, less than 51.76% of Python3 online submissions for Two Sum.

太嚇人，遍歷太慢了。

思路 2

A + B = target，同樣的：A = target - B，我們可以維護一個字典來記錄說訪問過的值。

這裡我犯了個錯，我先貼程式碼

class Solution:
def twoSum(self, nums: List[int], target: int)-> List[int]:
lookup = {}
for i, j in enumerate(nums):
if target - j not in lookup.values():
lookup[i] = j
else:
return [i, nums.index(target - i)]

Runtime: 568 ms, faster than 34.25% of Python3 online submissions forTwo Sum.
Memory Usage: 14.1 MB, less than 9.73% of Python3 online submissions for Two Sum.

雖然比第一種解法好，但是才超過34.25%，這也太慢了吧。

其實是我寫法不夠好，字典維序反了，這樣查詢會慢很多，每次查詢都需要查詢 lookup.values() ，這其實是一個列表，應該就是這裡導致速度變慢。

改良後的程式碼：

class Solution:
def twoSum(self, nums: List[int], target: int)-> List[int]:
lookup = {}
for i, j in enumerate(nums):
if target - j in lookup:
return [i, nums.index(target - j)]
else:
lookup[j] = i

Runtime: 40 ms, faster than 73.58% of Python3 online submissions for Two Sum.
Memory Usage: 14.1 MB, less than 10.10% of Python3 online submissions for Two Sum.

比73.58%的要快，果然細節決定成敗啊。