https://leetcode.com/problems/valid-anagram/description/

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.

Follow up:

What if the inputs contain unicode characters? How would you adapt your solution to such case?

• 字符串处理，频率统计题。
• 第一种方法是对字符串按字母排序，然后比较是否相等，简洁明了。
• 第二种方法是用hash table。此题指明只有小写字母，所以‘a'-'z'26个英文字母，做张表就可以了。
• 注意判断方法有两种，一种是都统计完再判断，一种是对s做加法，对t做减法，同时做减法时如果判断有负数统计出现，说明t中出现了多余字母时s中没有的。
• https://leetcode.com/problems/valid-anagram/solution/

1 // 2 //  main.cpp 3 //  LeetCode 4 // 5 //  Created by Hao on 2017/3/16. 6 //  Copyright © 2017年 Hao. All rights reserved. 7 // 8  9 #include 
       
        10 #include 
        
         11 #include 
         
          12 using namespace std;13 14 class Solution {15 public:16     // STL sort17     bool isAnagram(string s, string t) {18         sort(s.begin(), s.end());19         sort(t.begin(), t.end());20         21         return s == t;22     }23     24     // Hash25     bool isAnagram2(string s, string t) {26         if (s.length() != t.length()) {27             return false;28         }29         30         vector
          
            dict(26);31         32         for (auto ch : s) {33             ++ dict[ch - 'a'];34         }35         36         for (auto ch : t) {37             -- dict[ch - 'a'];38         }39         40         for (auto iter : dict) {41             if (iter != 0)42                 return false;43         }44         45         return true;46     }47     48     bool isAnagram3(string s, string t) {49         if (s.length() != t.length()) {50             return false;51         }52         53         vector
           
             dict(26);54         55         for (auto ch : s) {56             ++ dict[ch - 'a'];57         }58         59         for (auto ch : t) {60             -- dict[ch - 'a'];61             62             if (dict[ch - 'a'] < 0) // Extra characters in string t63                 return false;64         }65         66         return true;67     }68 };69 70 int main(int argc, char* argv[])71 {72     Solution    testSolution;73     int result = 1;74     75     result = result && testSolution.isAnagram("anagram", "nagaram");76     result = result && testSolution.isAnagram("program", "margorp");77     result = result && !testSolution.isAnagram("rat", "car");78     result = result && !testSolution.isAnagram("aaab", "aaa");79     result = result && testSolution.isAnagram("", "");80     81     result = result && testSolution.isAnagram2("anagram", "nagaram");82     result = result && testSolution.isAnagram2("program", "margorp");83     result = result && !testSolution.isAnagram2("rat", "car");84     result = result && !testSolution.isAnagram2("aaab", "aaa");85     result = result && testSolution.isAnagram2("", "");86 87     result = result && testSolution.isAnagram3("anagram", "nagaram");88     result = result && testSolution.isAnagram3("program", "margorp");89     result = result && !testSolution.isAnagram3("rat", "car");90     result = result && !testSolution.isAnagram3("aaab", "aaa");91     result = result && testSolution.isAnagram3("", "");92 93     if (result)94         cout << "ALl tests pass!" << endl;95     else96         cout << "Tests fail!" << endl;97     98     return 0;99 }