[leetcode]242. Valid Anagram

Given two strings s and t, write a function to determine if t is an anagram of s.

For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.

Note:
You may assume the string contains only lowercase alphabets.

判断的你两个字符串是否由相同的字符组成

解法1:

排序,时间复杂度O(N log N)

1 class Solution(object):
2     def isAnagram(self, s, t):
3         return sorted(s) == sorted(t)
4         

解法2:

用字典统计s中每个字母出现的次数,然后对照t中字母出现的次数,不匹配返回False,O(N)

1 class Solution(object):
2     def isAnagram(self, s, t):
3         d = collections.Counter(s)
4         for c in t:
5             d[c] -= 1
6         for val in d.values():
7             if val != 0:
8                 return False
9         return True

 C语言:

 1 bool isAnagram(char* s, char* t) {
 2     int len_s,len_t;
 3     len_s=strlen(s);
 4     len_t=strlen(t);
 5     if(len_s!=len_t) return false;
 6     int *flag = malloc(sizeof(int)*26);
 7     for(int i=0;i<26;i++)
 8         flag[i] = 0;
 9     for(int i=0;i<len_s;i++){
10         flag[s[i]-'a'] += 1;
11         flag[t[i]-'a'] -= 1;
12     }
13     for(int i=0;i<26;i++)
14         if(flag[i] != 0) {
15             free(flag);
16             return false;
17         }
18     free(flag);
19     return true;
20 }
原文地址:https://www.cnblogs.com/fcyworld/p/6506880.html