查找排序算法联系题

一 给定两个字符串s和t，判断t是否为s的重新排列后组成的单词

• s = "anagram", t = "nagaram", return True
• s = "rat", t = "car", return False
• leetcode地址：https://leetcode.com/problems/valid-anagram/description/

解法一：排序，O(n*logn)

class Solution:
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        ss = list(s)
        tt = list(t)
        ss.sort()
        tt.sort()
        return ss == tt
"""
输入："anagram"、"nagaram"
输出：true
Runtime: 32 ms
"""

排序方法简写如下：

class Solution:
    def isAnagram(self, s, t):
        return sorted(list(s)) == sorted(list(t))

解法二：判断字母数量一致，时间复杂度O(n)

class Solution:
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        dict1 = {}   # 用字典来维护字符的数量
        dict2 = {}
        for ch in s:
            dict1[ch] = dict1.get(ch, 0) + 1   # 没有就新建，有就加1
        for ch in t:
            dict2[ch] = dict2.get(ch, 0) + 1
        return dict1 == dict2
 
"""
输入："anagram"，"nagaram"
输出：true
Runtime: 32 ms
"""