1. 题目
1.1 题目解析
- T242. 有效的字母异位词 国际站讨论
由于题目想对比的两个字符串长度一致,故我们转换成ASCII码,然后进行比较。
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
return abs(sum(ord(i)**0.5 for i in s) - sum(ord(j)**0.5 for j in t) )< 1e-5
时间复杂度O(n)
- T49. 字母异位词分组
class Solution:
def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
d = {}
for w in sorted(strs):
key = tuple(sorted(w))
d[key] = d.get(key,[]) + [w]
return sorted(d.values())
时间复杂度O(nlogn)
- T1. 两数之和
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
hash = {}
for ind ,num in enumerate(nums):
hash[num] = ind
for i, num in enumerate(nums):
j = hash.get(target - num)
if j != None and j != i:
return [i,j]
时间复杂度O(n)