[LeetCode] 242. Valid Anagram 验证变位词

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?

给2个字符串s和t，写一个判断二者是否为变位词的函数。

解法1：利用HashMap记录s中出现的字母和数量，然后在用t的字母和数量来验证。

解法2：把两个string变为char array，对两个array进行排序，然后比较是否一样。

Java：

class Solution {
    public boolean isAnagram(String s, String t) {
        HashMap<Character, Integer> map = new HashMap<>();
        
        // first time: store each s char and occurrence into map
        for(int i=0; i<s.length(); i++) {
            char sChar = s.charAt(i);
            map.put(sChar, map.getOrDefault(sChar, 0) + 1);
        }
        // second time: compare t char with map to see match or not
        for(int i=0; i<t.length(); i++) {
            char tChar = t.charAt(i);
            
            if(!map.containsKey(tChar))
                return false;
            
            if(map.get(tChar) == 1)
                map.remove(tChar);
            else
                map.put(tChar, map.get(tChar) - 1);
                       
        }
        
        return map.size() == 0 ? true : false;
    }
}　

Java:

public class Solution {
    public boolean isAnagram(String s, String t) {
        int[] alphabet = new int[26];
        for (int i = 0; i < s.length(); i++) alphabet[s.charAt(i) - 'a']++;
        for (int i = 0; i < t.length(); i++) alphabet[t.charAt(i) - 'a']--;
        for (int i : alphabet) if (i != 0) return false;
        return true;
    }
}　　

Python:  T: O(n)  S: O(1)

class Solution:
    def isAnagram(self, s, t):
        if len(s) != len(t):
            return False

        count = {}

        for c in s:
            if c.lower() in count:
                count[c.lower()] += 1
            else:
                count[c.lower()] = 1

        for c in t:
            if c.lower() in count:
                count[c.lower()] -= 1
            else:
                count[c.lower()] = -1
            if count[c.lower()] < 0:
                return False

        return True

Python: wo

class Solution(object):
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        a = [0] * 256
        for i in s:
            a[ord(i)] += 1
            
        b = [0] * 256    
        for j in t:
            b[ord(j)] += 1
            
        return  a == b   　　

Python:  T: O(n)  S: O(1)

class Solution:
    def isAnagram3(self, s, t):
        if len(s) != len(t):
            return False 
        count = collections.defaultdict(int)
        for c in s:
            count[c] += 1
        for c in t:
            count[c] -= 1
            if count[c] < 0:
                return False
        return True

Python:

def isAnagram1(self, s, t):
    dic1, dic2 = {}, {}
    for item in s:
        dic1[item] = dic1.get(item, 0) + 1
    for item in t:
        dic2[item] = dic2.get(item, 0) + 1
    return dic1 == dic2　　

Python:  T: O(nlogn)  S: O(n)

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

C++:

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

　　　　

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