题目
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?
思路1:排序
将两个字符串分别转换为字符数组并调用Arrays.sort()方法对字符数组进行排序,并判断排序后的数组是否相同
时间复杂度:涉及到排序,为O(nlgn)
代码
class Solution {
public boolean isAnagram(String s, String t) {
if (s.length() != t.length()) {
return false;
}
char[] str1 = s.toCharArray();
char[] str2 = t.toCharArray();
Arrays.sort(str1);
Arrays.sort(str2);
return Arrays.equals(str1, str2);
}
}
Runtime: 5 ms, faster than 56.20% of Java online submissions for Valid Anagram.
Memory Usage: 36.5 MB, less than 78.36% of Java online submissions for Valid Anagram.
思路2:hashtable
用重复检测的方法来考虑:
用一个数组来纪录字符串中各个字母出现的次数,并且为了节省空间可以只用一个数组完成,遍历一趟,s中的某字母出现一次就让对应位置计数+1,反之,t中字母出现时使计数-1。若t是s的"anagram",那么最后这个counter数组中的所有元素都应该是0。
unicode的情况把数组换成hashtable。
时间复杂度O(n)
代码
class Solution {
public boolean isAnagram(String s, String t) {
if (s.length() != t.length()) {
return false;
}
int[] counter = new int[26];//因为题目规定字符都是小写的,所以用26就能搞定
for(int i = 0; i < s.length(); i ++){
counter[s.charAt(i) - 'a']++; // - 'a' 用来找到相对a的偏移值,也就是找到对应计数的位置
counter[t.charAt(i) - 'a']--;
}
for(int count : counter ){
if(count != 0){
return false;
}
}
return true;
}
}
Runtime: 3 ms, faster than 89.10% of Java online submissions for Valid Anagram.
Memory Usage: 34.3 MB, less than 100.00% of Java online submissions for Valid Anagram.