Algorithm
本周选择的算法题是：Letter Combinations of a Phone Number
规则如下：
Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
Example:
Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
Solution
Runtime：24 ms，快过 91.28%。
Memory：13.6 MB，低于 5.88%。
class Solution:
map = {
'2': 'abc',
'3': 'def',
'4': 'ghi',
'5': 'jkl',
'6': 'mno',
'7': 'pqrs',
'8': 'tuv',
'9': 'wxyz',
}
def letterCombinations(self, digits: str) -> List[str]:
ans = []
def recursive(letters, remain):
if len(remain) == 0:
ans.append(letters)
else:
for letter in Solution.map[remain[0]]:
recursive(letters + letter, remain[1:])
if digits: recursive('', digits)
return ans
基于栈的 BFS 算法：
class Solution:
map = {
'2': 'abc',
'3': 'def',
'4': 'ghi',
'5': 'jkl',
'6': 'mno',
'7': 'pqrs',
'8': 'tuv',
'9': 'wxyz',
}
def letterCombinations(self, digits: str) -> List[str]:
if not digits: return []
ans = [""]
while ans and len(ans[0]) != len(digits):
letter = ans.pop(0)
for char in Solution.map[digits[len(letter)]]:
ans.append(letter + char)
return ans
还可以将字典优化成数组，减少内存的的占用：
# map = {
# '2': 'abc',
# '3': 'def',
# '4': 'ghi',
# '5': 'jkl',
# '6': 'mno',
# '7': 'pqrs',
# '8': 'tuv',
# '9': 'wxyz',
# }
map = [ 'abc','def','ghi','jkl','mno','pqrs','tuv','wxyz' ]
Review
How to create your own pull to refresh / custom refresh indicator widget in Flutter. 一个 Flutter 下拉刷新 widget 的介绍，可以感受下和 iOS 控件开发体验有何不同。
Tip
在 LLDB 中对所有的 +load 方法添加断点：
br s -r "\+\[.+ load\]$"
Share
学习与认知，是有复利效应的，很多内容都可以触类旁通。
我们能做到的就是，不断丰富自己的知识和思想体系，慢慢积累，说不定，哪天就用到了。当然，现在大家的生活节奏都很快，更需要计划性和体系化。