49 - Group Anagrams

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.

思路解析：

1、先将向量中的字符串按字典排序（因为必须按字典的顺序，因此先排序）

也就是[ate, bat, eat, nat, tan, tea]

2、将向量中的每个字符串排序，并且将其中字典序最小的作为map中的键值

aet -> ate,eat,tea

bat -> bat,

ant->nat,tan

/*

*/

#include "stdafx.h"
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>

using namespace std;

class Solution_049_GroupAnagrams
{
public:
	vector<vector<string>> groupAnagrams(vector<string>& strs) 
	{
		if (strs.empty())
		{
			return vector<vector<string> >();
		}
		
		int len = strs.size();

		//将字符串数组按照字典顺序排列
		sort(strs.begin(), strs.end());

		vector<vector<string> > ret;

		//利用哈希思想构建map，将排序后相等的字符串存在相应的vector
		map<string, vector<string> > mv;

		for (int i = 0; i < len; i++)
		{
			string str = strs[i];
			sort(str.begin(), str.end());

			//每个相同的str都有一个向量与之对应
			mv[str].push_back(strs[i]);
		}

		for (map<string, vector<string>>::iterator iter = mv.begin(); iter != mv.end(); iter++)
		{
			ret.push_back(iter->second);
		}

		return ret;
	}
};