LintCode 171. Anagrams

171. Anagrams
     中文English
     Given an array of strings, return all groups of strings that are anagrams.

Example
Example 1:

Input:[“lint”, “intl”, “inlt”, “code”]
Output:[“lint”, “inlt”, “intl”]
Example 2:

Input:[“ab”, “ba”, “cd”, “dc”, “e”]
Output: [“ab”, “ba”, “cd”, “dc”]
Challenge
What is Anagram?

Two strings are anagram if they can be the same after change the order of characters.
Notice
All inputs will be in lower-case

解法1：用map。
注意只有map才能用for(auto m : mp)，unordered_map不能用，因为unordered_map是基于hash table，无序，无法遍历。
补充：好像不对。因为LintCode 78 "Longest Common Prefix"那题我就是用的auto遍历unordered_map。这里不知道为什么。下次想想。

代码如下：

class Solution {
public:
    /**
     * @param strs: A list of strings
     * @return: A list of strings
     */
    vector<string> anagrams(vector<string> &strs) {
        int N = strs.size();
        if (N == 0) return vector<string>();
        
        map<vector<int>, vector<string>> mp;
        vector<string> result;
        
        for (int i = 0; i < N; ++i) {
            vector<int> charMap(26, 0);
            int len = strs[i].size();
            for (int j = 0; j < len; ++j) {
                charMap[strs[i][j] - 'a']++;
            }
            mp[charMap].push_back(strs[i]);
        }
        
        for (auto m : mp) {
            if (m.second.size() > 1) {
                for (auto s : m.second) result.push_back(s);
            }
        }
        
        return result;
    }

};

注意：上面说的unordered_map不能用于用for(auto m : mp)是错的！这里根本原因是因为如果我们把vector作为unordered_map的key的话，unordered_map不知道对于vector<>的hash函数怎么生成。所以我们要自己创建一个对于vector<>的hash函数。下面这个帖子讲的不错！
https://jimmy-shen.medium.com/stl-map-unordered-map-with-a-vector-for-the-key-f30e5f670bae

struct VectorHasher {
    int operator()(const vector<int> &V) const {
        int hash = V.size();
        for(auto &i : V) {
            hash ^= i + 0x9e3779b9 + (hash << 6) + (hash >> 2);
        }
        return hash;
    }
};

class Solution {
public:
    /**
     * @param strs: A list of strings
     * @return: A list of strings
     */

    vector<string> anagrams(vector<string> &strs) {
        int N = strs.size();
        if (N == 0) return vector<string>();
        
        unordered_map<vector<int>, vector<string>, VectorHasher> mp;
        vector<string> result;
        
        for (int i = 0; i < N; ++i) {
            vector<int> charMap(26, 0);
            int len = strs[i].size();
            for (int j = 0; j < len; ++j) {
                charMap[strs[i][j] - 'a']++;
            }
            mp[charMap].push_back(strs[i]);
        }
        
        for (auto m : mp) {
            if (m.second.size() > 1) {
                for (auto s : m.second) result.push_back(s);
            }
        }
        return result;
    }
};

二刷：用unordered_map，这里我们用string，所以不存在hash函数找不到的问题。

class Solution {
public:
    /**
     * @param strs: A list of strings
     * @return: A list of strings
     */
    vector<string> anagrams(vector<string> &strs) {
        int n = strs.size();
        unordered_map<string, vector<string>> um;
        vector<string> res;
        for (auto str : strs) {
            string tmpStr = str;
            sort(tmpStr.begin(), tmpStr.end());
            um[tmpStr].push_back(str);
        }
        for (auto m : um) {
            if (m.second.size() > 1) {
                res.insert(res.end(), m.second.begin(), m.second.end());
            }
        }
        return res;
    }
};