LeetCode 49. Group Anagrams

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本文介绍了一种使用哈希映射实现的高效算法来解决字谜游戏问题，即对一组字符串进行分组，将互为字谜(anagrams)的字符串归类到一起。文章通过具体的代码示例展示了如何利用C++语言中的`unordered_map`数据结构来实现这一算法，并确保输出结果遵循字典序排列。

Given an array of strings, group anagrams together.

For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],
Return:

[
  ["ate", "eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Note:

For the return value, each inner list's elements must follow the lexicographic order.
All inputs will be in lower-case.

// Usage of HashMap.

#include <string>
#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;

vector< vector<string> > groupAnagrams(vector<string>& strs) {
    if(strs.size() == 0) return {{}};
    vector< vector<string> > res;
    unordered_map<string, vector<string> > map;
    for(int i = 0; i < strs.size(); ++i) {
        string tmp = strs[i];
        sort(tmp.begin(), tmp.end());   // this is the key point.
        auto iter = map.find(tmp);
        if(iter == map.end()) {
            map.insert({tmp, {strs[i]}});  // make_pair {}
        } else {
            (iter->second).push_back(strs[i]);
        }
    }
    auto it = map.begin();
    while(it != map.end()) {
        sort(it->second.begin(), it->second.end());  // sort them in lexicographic order.
        res.push_back(it->second);
        it++;
    }
    return res;
}

int main(void) {
    vector<string> inputs{"eat", "tea", "tan", "ate", "nat", "bat"};
    vector< vector<string> > res = groupAnagrams(inputs);
    for(int i = 0; i < res.size(); ++i) {
        for(int j = 0; j < res[i].size(); ++j) {
            cout << res[i][j] << endl;
        }
        cout << endl;
    }
    cout << endl;
}