(self, s1, s2)
| 24 | # 计算每个字母出现的次数并存入到相应的list中 |
| 25 | # 比较两个list是否相同 |
| 26 | def anagramSolution2(self, s1, s2): |
| 27 | c1 = [0] * 26 |
| 28 | c2 = [0] * 26 |
| 29 | |
| 30 | for i in range(len(s1)): |
| 31 | pos = ord(s1[i]) - ord('a') |
| 32 | c1[pos] = c1[pos] + 1 |
| 33 | |
| 34 | for i in range(len(s2)): |
| 35 | pos = ord(s2[i]) - ord('a') |
| 36 | c2[pos] = c2[pos] + 1 |
| 37 | |
| 38 | j = 0 |
| 39 | stillOK = True |
| 40 | while j < 26 and stillOK: |
| 41 | if c1[j] == c2[j]: |
| 42 | j = j + 1 |
| 43 | else: |
| 44 | stillOK = False |
| 45 | |
| 46 | return stillOK |
| 47 | |
| 48 | # 首先将两个字符串list化 |
| 49 | # 将两个list中的字符生成两个set |