242. Valid Anagram

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

Example 1:

Input: s = “anagram”, t = “nagaram”
Output: true

Example 2:

Input: s = “rat”, t = “car”
Output: 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?

Solution1:

使用sort

class Solution {
 public:
  bool isAnagram(string s, string t) {
    if (s.length() != t.length()) {
      return false;
    }
    sort(s.begin(), s.end());
    sort(t.begin(), t.end());

    for (int i = 0; i < s.length(); i++) {
      if (s[i] != t[i]) return false;
    }
    return true;
  }
};

简化版本

class Solution {
 public:
  bool isAnagram(string s, string t) {
    sort(s.begin(), s.end());
    sort(t.begin(), t.end());
    if (s == t) return true;
    return false;
  }
};

Solution2:

建立Hash表, key是char, 对应该字符出现的次数. 一个字符串加一个字符串减。
这里使用散列表实现的unodered_map

class Solution {
 public:
  bool isAnagram(string s, string t) {
    if (s.length() != t.length()) return false;
    unordered_map<char, int> counts;
    for (int i = 0; i < s.length(); i++) {
      counts[s[i]]++;
      counts[t[i]]--;
    }
    for (auto count : counts) {
      if (count.second) return false;
    }
    return true;
  }
};

Solution3

用ascii码作为key值, 不用map容器用数组

class Solution {
 public:
  bool isAnagram(string s, string t) {
    if (s.length() != t.length()) return false;
    int counts[26];
    memset(counts, 0, sizeof(counts));
    for (int i = 0; i < s.length(); i++) {
      counts[s[i] - 'a']++;
      counts[t[i] - 'a']--;
    }
    for (int i = 0; i < 26; i++) {
      if (counts[i]) return false;
    }
    return true;
  }
};