leetcode 242. Valid Anagram | Map

# Description

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.

# My solution

基本思路是利用dict思想, 统计每个单词出现次数, 最后对两次统计的进行对照.

```

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

```

# Discuss

基本思想一致, 但是更节省内存的一种方式. 有点像多数投票问题中的解决方案, 在同一个统计数组++--, 最后判断是否每个位置为0即可.

```

class Solution {public:    bool isAnagram(string s, string t) {        if (s.length() != t.length()) return false;        int n = s.length();        int counts[26] = {0};        for (int i = 0; i < n; i++) {             counts[s[i] - 'a']++;            counts[t[i] - 'a']--;        }        for (int i = 0; i < 26; i++)            if (counts[i]) return false;        return true;    }};

```

注意到c++中map的用法(对比于python dict), 网友的一种方案如下:

```

class Solution {public:    bool isAnagram(string s, string t) {        if (s.length() != t.length()) return false;        int n = s.length();        unordered_map<char, int> counts;        for (int i = 0; i < n; i++) {            counts[s[i]]++;            counts[t[i]]--;        }        for (auto count : counts)            if (count.second) return false;        return true;    }};

```

# Reference

- [leetcode 242](https://leetcode.com/problems/valid-anagram/description/)

- [std::unordered_map](http://www.cplusplus.com/reference/unordered_map/unordered_map/)