Exercise of the Day (44): Effective Alphabetical Words

---

title: Exercise of the Day (44): Valid Alphabets

categories:[Swords offer]

tags:[Daily practice]

date: 2022/04/18

---

Exercise of the Day (44): Effective Alphabetical Words

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

Note: If each character in s and t appears the same number of times, then s and t are called anagrams of each other.

Example 1:

Input: s = "anagram", t = "nagaram"

output: true

Example 2:

Input: s = "rat", t = "car"

output: false

hint:

1 <= s.length, t.length <= 5 * 104

s and t contain only lowercase letters

Source: LeetCode

Link: https://leetcode-cn.com/problems/valid-anagram

Method 1: Sort

Thought analysis

t is an anagram of s equivalent to "two strings sorted equal". Therefore, we can sort the strings s and t respectively, and judge whether the sorted strings are equal. Furthermore, if the lengths of s and t

Different, t is not necessarily an anagram of s.

 bool isAnagram(string s, string t) {
    if (s.length() != t.length()) {
        return false;
    }
    sort(s.begin(), s.end());
    sort(t.begin(), t.end());
    return s == t;
}

Method 2: Hash table

Thought analysis

Since the string contains only 26 lowercase letters, we can maintain a frequency array table with a length of 26, first traverse the frequency of the characters in the record string s, and then traverse the characters

String t, minus the corresponding frequency in table, if table[i]<0, it means that tt contains an extra character that is not in s, just return false

 bool isAnagram(string s, string t) {
    if (s.length() != t.length()) {
        return false;
    }
    vector<int> table(26, 0);
    for (auto &ch : s) {
        table[ch - 'a']++;
    }
    for (auto &ch : t) {
        table[ch - 'a']--;
        if (table[ch - 'a'] < 0) {
            return false;
        }
    }
    return true;
}