题解 | 字母异位词的长度

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param s string字符串 
# @param c string字符串 
# @return int整型
#
class Solution:
    def isCongruent(self , s: str, c: str) -> int:
        # write code here
        # dic = {}
        arr1 = [[0,0] for _ in range(26)]
        arr2 = [[0,0] for _ in range(26)]

        for i in range(len(s)):
            arr1[ord(s[i])-ord('a')][1] += 1
        for j in range(len(c)):
            arr2[ord(c[j])-ord('a')][1] += 1
            
        for i in range(26):
            if arr1[i][1] != arr2[i][1]:
                return -1
        return len(s)