今天選題的方式 不是按序 也不是偷懶拿之前寫的(會標記Review) 是 Pick One (應該是隨機選題吧) <- 如上圖

Description

Given two strings s and t , write a function to determine if t is an anagram of s.

Example 1:

Input: s = "anagram", t = "nagaram"
Output: true
Example 2:

Input: s = "rat", t = "car"
Output: false
Note:
You may assume the string contains only lowercase alphabets.

Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?

思路

英文必須好

anagram == 重組

整句翻下來是 字串t 是否為 字串s的重組

而 解法是 長度一樣 , 把字串放進list後，進行sort，再繼續判斷是否兩個list slist ，tlist 是否一模模沒兩樣，是則 回傳 True
不一樣 直接 回傳 False

正解

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        slist = []
        tlist = []
        if len(s)!=len(t):
            return False
        for i in range(len(s)):
            slist.append(s[i])
            tlist.append(t[i])
        
        slist.sort()    
        tlist.sort()
        
        for j in range(len(s)):
            if slist[j]!=tlist[j]:
                return False
        
        return True

Result