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83 lines (71 loc) · 2.33 KB
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// Given two strings s and t , write a function to determine if t is an anagram of s.
// See: https://leetcode.com/problems/valid-anagram/
package leetcode.array_and_hashtable;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
public class ValidAnagram {
/**
* Solution 4 - Array solution: O(n) time, O(1) space (Fast solution).
*/
public boolean isAnagram(String s, String t) {
if (s.length() != t.length()) return false;
int[] alphabet = new int[26];
char[] ch1 = s.toCharArray();
char[] ch2 = t.toCharArray();
for (int i = 0; i < ch1.length; i++) {
alphabet[ch1[i] - 'a']++;
alphabet[ch2[i] - 'a']--;
}
for (int i : alphabet)
if (i != 0)
return false;
return true;
}
/**
* Solution 3 - Sort solution. O(n.log(n)) time, O(1) space.
*/
public boolean isAnagram_var3(String s, String t) {
if (s.length() != t.length()) return false;
char[] s1 = s.toCharArray();
char[] s2 = t.toCharArray();
Arrays.sort(s1);
Arrays.sort(s2);
return Arrays.equals(s1, s2);
}
/**
* Solution 2 - HashMap solution.
*/
public boolean isAnagram_var2(String s, String t) {
if (s.length() != t.length()) return false;
Map<Character, Integer> map = new HashMap<>();
for (char ch : t.toCharArray())
map.put(ch, map.getOrDefault(ch, 0) + 1);
for (char ch : s.toCharArray()) {
if (!map.containsKey(ch)) {
return false;
} else {
int count = map.get(ch) - 1;
if (count == 0) {
map.remove(ch);
} else {
map.put(ch, count);
}
}
}
return true;
}
/**
* Solution 1 - Array solution.
*/
public boolean isAnagram_var1(String s, String t) {
if (s.length() != t.length()) return false;
int[] alphabet1 = new int[26];
int[] alphabet2 = new int[26];
for (int ch : s.toCharArray())
alphabet1[ch - 'a']++;
for (int ch : t.toCharArray())
alphabet2[ch - 'a']++;
return Arrays.equals(alphabet1, alphabet2);
}
}