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:

  1. Input: s = "anagram", t = "nagaram"
  2. Output: true

Example 2:

  1. Input: s = "rat", t = "car"
  2. 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?

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。

代码实现

Java

Hash

  1. class Solution {
  2.     public boolean isAnagram(String s, String t) {
  3.         if (s.length() != t.length()) {
  4.             return false;
  5.         }
  7.         int count[] = new int[26];
  9.         for (int i = 0; i < s.length(); i++) {
  10.             count[s.charAt(i) - 'a']++;
  11.             count[t.charAt(i) - 'a']--;
  12.         }
  14.         for (int i = 0; i < 26; i++) {
  15.             if (count[i] != 0) {
  16.                 return false;
  17.             }
  18.         }
  20.         return true;
  21.     }
  22. }

排序

  1. class Solution {
  2.     public boolean isAnagram(String s, String t) {
  3.         if (s.length() != t.length()) {
  4.             return false;
  5.         }
  7.         char[] ss = s.toCharArray();
  8.         char[] tt = t.toCharArray();
  9.         Arrays.sort(ss);
  10.         Arrays.sort(tt);
  12.         for (int i = 0; i < ss.length; i++) {
  13.             if (ss[i] != tt[i]) {
  14.                 return false;
  15.             }
  16.         }
  18.         return true;
  19.     }
  20. }

Follow up

  1. class Solution {
  2.     public boolean isAnagram(String s, String t) {
  3.         if (s.length() != t.length()) {
  4.             return false;
  5.         }
  7.         Map<Character, Integer> map = new HashMap<>();
  9.         for (int i = 0; i < s.length(); i++) {
  10.             char c = s.charAt(i);
  11.             if (!map.containsKey(c)) {
  12.                 map.put(c, 1);
  13.             } else {
  14.                 map.put(c, map.get(c) + 1);
  15.             }
  16.         }
  18.         for (int i = 0; i < t.length(); i++) {
  19.             char c = t.charAt(i);
  20.             if (!map.containsKey(c) || map.get(c) == 0) {
  21.                 return false;
  22.             } else {
  23.                 map.put(c, map.get(c) - 1);
  24.             }
  25.         }
  27.         return true;
  28.     }
  29. }

JavaScript

  1. /**
  2.  * @param {string} s
  3.  * @param {string} t
  4.  * @return {boolean}
  5.  */
  6. var isAnagram = function (s, t) {
  7.   if (s.length !== t.length) return false
  8.   const cnt = new Map()
  9.   for (const c of s) {
  10.     if (!cnt.has(c)) cnt.set(c, 1)
  11.     else cnt.set(c, cnt.get(c) + 1)
  12.   }
  13.   for (const c of t) {
  14.     if (cnt.has(c) && cnt.get(c) > 0) cnt.set(c, cnt.get(c) - 1)
  15.     else return false
  16.   }
  17.   return true
  18. }