leetcode解题之242# Valid Anagram Java版 （判断两个字母串是否具有相同的字母组合）
16lz
2021-01-22
242. Valid Anagram
For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.
Note:
You may assume the string contains only lowercase alphabets.
What if the inputs contain unicode characters? How would you adapt your solution to such case
排序之后是否相等
import java.util.Arrays;
public class Solution {
public boolean isAnagram(String s, String t) {
char[] sArr = s.toCharArray();
char[] tArr = t.toCharArray();
Arrays.sort(sArr);
Arrays.sort(tArr);
return String.valueOf(sArr).equals(String.valueOf(tArr));
}
}
才用map，若map中含有此key，则value+1；否则 value置1，取的时候最后看是否为0，如果存在不为0的value，返回false
import java.util.HashMap;
import java.util.Map;
public class Solution {
public boolean isAnagram(String s, String t) {
//长度不相等，返回false
if (s.length() != t.length())
return false;
//key为出现的字母，value为字母出现的次数
Map<Character, Integer> map = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
if (map.containsKey(s.charAt(i))) {
map.put(s.charAt(i), map.get(s.charAt(i)) + 1);
} else
map.put(s.charAt(i), 1);
}
for (int i = 0; i < t.length(); i++) {
if (map.containsKey(t.charAt(i))) {
map.put(t.charAt(i), map.get(t.charAt(i)) - 1);
} else
return false;
}
for (int i = 0; i < s.length(); i++) {
if (map.get(s.charAt(i)) != 0)
return false;
}
return true;
}
}