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： 采用字典进行存储，对每个字母进行相应的统计
class Solution(object):
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
sn = len(s)
tn = len(t)
lists = []
listt = []
if sn != tn:
return False
for i in range(0, sn):
lists.append(s[i])
#print(lists)
#print("-----------------------------------")
for j in range(0, tn):
listt.append(t[j])
#print(listt)
listt.sort()
#print(listt)
lists.sort()
#print(lists)
if lists == listt:
return True
else:
return False
# letters = {}
# letters2 = {}
# for c in s:
# letters[c] = letters[c] + 1 if c in letters else 1
# #print(letters[c])
# print(letters)
# print("=======================================")
# for d in s:
# letters2[d] = letters2[d] + 1 if d in letters2 else 1
# #print(letters2[c])
# print(letters2)
# if letters == letters2:
# return True
# else:
# return False
if __name__ == "__main__":
a = "anagram"
b = "nagaram"
print(Solution().isAnagram(a, b))
另解：
def isAnagram2(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
"""
思路：定义一个26长度的列表，每个元素都是0，在s中进行扫描，扫描到，对应列表元素位置加一
在t中扫描，相应元素再减一。
"""
a = [0] * 26
for c in s:
a[ord(c)-ord('a')] += 1
for c in t:
a[ord(c)-ord('a')] -= 1
for sc in a:
if sc != 0:
return False
return True
