(self, s1, s2)
| 3 | # 对字符串对应的两个list进行排序 |
| 4 | # 依次比较字符是否匹配 |
| 5 | def anagramSolution1(self, s1, s2): |
| 6 | alist1 = list(s1) |
| 7 | alist2 = list(s2) |
| 8 | |
| 9 | alist1.sort() |
| 10 | alist2.sort() |
| 11 | |
| 12 | pos = 0 |
| 13 | matches = True |
| 14 | |
| 15 | while pos < len(s1) and matches: |
| 16 | if alist1[pos] == alist2[pos]: |
| 17 | pos = pos + 1 |
| 18 | else: |
| 19 | matches = False |
| 20 | |
| 21 | return matches |
| 22 | |
| 23 | # 首先生成两个26个字母的list |
| 24 | # 计算每个字母出现的次数并存入到相应的list中 |