LeetCode Valid Anagram

LeetCode Valid Anagram

题目

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.

Follow up: What if the inputs contain unicode characters? How would
you adapt your solution to such case?

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思路&代码

对于只有小写字母的情况，只需要开一个26个int大小的数组对两个字符串进行字母统计即可，对一个字符串加而对另一个字符串减。这样，如果最后统计数组有一个字符次数不为0，即可知道不一样。

class Solution {public:    bool isAnagram(string s, string t) {        int charNums[26];        memset(charNums, 0, sizeof(charNums));        for (int i = s.size() - 1; i >= 0; i--) charNums[s[i] - 'a']++;        for (int i = t.size() - 1; i >= 0; i--) {            charNums[t[i] - 'a']--;            if (charNums[t[i] - 'a'] < 0) return false;        }        for (int i = 0; i < 26; i++) if (charNums[i]) return false;        return true;    }};

但是，如果字符有可能包含Unicode字符，也就是有65536个字符的情况下，当然我们思路也是一样，也可以使用数组来统计。但是这样就会造成极大的空间浪费，所以我选择牺牲时间而节约空间使用map来统计，时间上大概增加了100ms（仅在此题中）。

class Solution {public:    bool isAnagram(string s, string t) {        map<char, int> hash;        for (int i = s.size() - 1; i >= 0; i--) {            if (hash.find(s[i]) == hash.end()) hash[s[i]] = 1;            else hash[s[i]]++;        }        for (int i = t.size() - 1; i >= 0; i--) {            if (hash.find(t[i]) == hash.end()) return false;            else hash[t[i]]--;        }        for (map<char, int>::iterator iter = hash.begin(); iter != hash.end(); iter++)             if (iter->second != 0) return false;        return true;    } };