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?
Solutions
1. Dict
# Time: O(n)
# Space: O(n)
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
if not s and not t:
return True
if not s or not t or len(s) != len(t):
return False
sd = self.to_dict(s)
td = self.to_dict(t)
return sd == td
def to_dict(self, s):
if not s:
return {}
dic = {}
for c in s:
dic[c] = dic.get(c, 0) + 1
return dic
# 34/34 cases passed (52 ms)
# Your runtime beats 60.73 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (13.1 MB)
2. Array
public class Solution {
public boolean isAnagram(String s, String t) {
int[] table = new int[26];
// 记录字母在第一个单词中出现的次数
for(int i = 0; i < s.length(); i++){
table[s.charAt(i)-'a']++;
}
// 减去字母在第二个单词中出现的次数
for(int i = 0; i < t.length(); i++){
table[t.charAt(i)-'a']--;
}
// 如果某个计数器不为0，则返回假
for(int i = 0; i < 26; i++){
if(table[i]!=0) return false;
}
return true;
}
}