Valid Anagram (E)

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?

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题意

判断两个给定的单词(只包含小写字母)是否互为anagram，即字母组成相同但排列顺序不同的单词。

思路

如果只包含小写字母，只需要遍历每一个字符串，统计每个字符串中每个字母出现的次数即可。

也可以将字符串排序后逐个比较。

Follow up仍可以用Hash的思想进行处理，只不过不能只开一个简单的数组来当hash表。

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代码实现

Java

Hash

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

        int count[] = new int[26];

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

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

排序

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

        char[] ss = s.toCharArray();
        char[] tt = t.toCharArray();
        Arrays.sort(ss);
        Arrays.sort(tt);
        
        for (int i = 0; i < ss.length; i++) {
            if (ss[i] != tt[i]) {
                return false;
            }
        }
        
        return true;
    }
}

Follow up

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

        Map<Character, Integer> map = new HashMap<>();

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (!map.containsKey(c)) {
                map.put(c, 1);
            } else {
                map.put(c, map.get(c) + 1);
            }
        }

        for (int i = 0; i < t.length(); i++) {
            char c = t.charAt(i);
            if (!map.containsKey(c) || map.get(c) == 0) {
                return false;
            } else {
                map.put(c, map.get(c) - 1);
            }
        }

        return true;
    }
}

JavaScript

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isAnagram = function (s, t) {
  if (s.length !== t.length) return false
  const cnt = new Map()
  for (const c of s) {
    if (!cnt.has(c)) cnt.set(c, 1)
    else cnt.set(c, cnt.get(c) + 1)
  }
  for (const c of t) {
    if (cnt.has(c) && cnt.get(c) > 0) cnt.set(c, cnt.get(c) - 1)
    else return false
  }
  return true
}