leetcode - Valid Anagram

最新推荐文章于 2018-11-02 16:42:07 发布

转载最新推荐文章于 2018-11-02 16:42:07 发布 · 60 阅读

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原文链接：http://www.cnblogs.com/shnj/p/4749216.html

本文介绍了一种用于判断两个字符串是否为字母异位词的有效算法。通过使用哈希表来记录每个字符的出现次数，并对比两个字符串中相同字符的出现次数是否一致，从而实现了快速准确的判断。

leetcode - Valid Anagram

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.

class Solution {
public:
    bool isAnagram(string s, string t) {
        if(s.length() != t.length()){
            return false;
        }
        map<char,int> sHash, tHash;
        string::iterator sit=s.begin(), tit=t.begin();
        while(sit!=s.end()){
            sHash[*sit]++;
            tHash[*tit]++;
            sit++;
            tit++;
        }
        map<char,int>::const_iterator sHashIt = sHash.begin();
        
        while(sHashIt != sHash.end()){
            map<char,int>::const_iterator tHashIt = tHash.find(sHashIt->first);
            if( tHashIt == tHash.end() || sHashIt->second != tHashIt->second)
            return false;
            sHashIt++;
        }
        return true;

    }
};