LeetCode OJ Valid Anagram 字符串

Valid Anagram Total Accepted: 8432 Total Submissions: 23731 My Submissions Question Solution

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

For example,

s = “anagram”, t = “nagaram”, return true.

s = “rat”, t = “car”, return false.

Note:

You may assume the string contains only lowercase alphabets.

意思就是比较两个字符串里的字符是否相同, 不关心字符的顺序

方法1 将两个字符串排序后, 比较是否相同

注意Java没有实现char[] toString() 方法, 需要用

String.valueOf()

将字符串数组转化为字符串.

[code]import java.util.Arrays;

public class ValidAnagram {
    public static void main(String[] args) {
        System.out.println(isAnagram("anagram", "anagram"));
        System.out.println(isAnagram("anagram", "nagaram"));
        System.out.println(isAnagram("car", "rat"));
        System.out.println(isAnagram("car", "rac"));
    }

    public static boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }

        char[] sChar = s.toCharArray();
        char[] tChar = t.toCharArray();

        Arrays.sort(sChar);
        Arrays.sort(tChar);

        return String.valueOf(sChar).equals(String.valueOf(tChar));

    }

}

方法2 比较每个字符的出现次数

新建26长度的数组存放每个字符出现的个数

注意可以用

s.charAt(i) - 'a'

来标记字符在数组中的位置

[code]import java.util.Arrays;

public class ValidAnagram {
    public static void main(String[] args) {
        System.out.println(isAnagram("anagram", "anagram"));
        System.out.println(isAnagram("anagram", "nagaram"));
        System.out.println(isAnagram("car", "rat"));
        System.out.println(isAnagram("car", "rac"));
    }

    public static boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }

        int[] counts = new int[26];

        for (int i = 0; i < s.length(); i++) {
            counts[s.charAt(i) - 'a']++;
        }
        for (int i = 0; i < t.length(); i++) {
            counts[t.charAt(i) - 'a']--;
        }

        for (int i = 0; i < 26; i++) {
            if (counts[i] != 0) {
                return false;
            }
        }

        return true;
    }

}

方法2 优化

注意到两个字符串长度不等时, 直接返回false.

长度相等时, 若两个字符串不是Anagram, 则counts[]其中肯定有负数, 可以提前在

for (int i = 0; i < t.length(); i++)

里进行判别.

问: counts[]在

for (int i = 0; i < t.length(); i++)

会先变成负数再变成正数吗? 不会, 因为

counts[t.charAt(i) - 'a']--

一直都是减减操作

[code]import java.util.Arrays;

public class ValidAnagram {
    public static void main(String[] args) {
        System.out.println(isAnagram("anagram", "anagram"));
        System.out.println(isAnagram("anagram", "nagaram"));
        System.out.println(isAnagram("car", "rat"));
        System.out.println(isAnagram("car", "rac"));
    }

    public static boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }

        int[] counts = new int[26];

        for (int i = 0; i < s.length(); i++) {
            counts[s.charAt(i) - 'a']++;
        }
        for (int i = 0; i < t.length(); i++) {
            counts[t.charAt(i) - 'a']--;
            if (counts[t.charAt(i) - 'a'] < 0) {
                return false;
            }
        }

        return true;
    }

}