map的一個例子

吧一個字典中，有換一個字母還是一個單詞並且超過15個的打出來。

#include<map>
#include<string>
#include<vector>
#include<iostream>
using namespace std;
/*
*map應用的簡單例子，說明了一些時候map怎麼工作
map<string,double>salaries;
salaries["Pat"]=75000;   //左邊調用operator[],因此插入“Pat”和一個值爲0的double到map，同屬返回這個double的引用，然後賦值將map中的double改爲75000
cout<<salaries["Pat"]<<endl;   //這個是可以輸出來的
cout<<salaries["Jan"]<<endl;   //這個不一定，要看程序，還有，玄學.....

map<string,double>::const_iterator itr;
itr=salaries.find("Chris");
if(itr==salaries.end())
	cout<<"Not ..."<<endl;
else
	cout<<itr->second<<endl;          //如果上面不是const_iterator,那麼這裏就可以給itr->second賦值了。

*/

/*《數據結構與算法剖析》,這本書裏的137頁例子，
找出來替換一個字母能形成15個以上單詞的單詞，
解決辦法就是用map，鍵存這個單詞，值是一個vectro，存替換一個字母形成的單詞們
*/

//構造了map後，直接檢查是不是至少15個，然後打印結果
void printHighChangeables(const map<string,vector<string>> & adjWords,int minWords=15);
void printHighChangeables(const map<string, vector<string>> & adjWords, int minWords = 15)
{
	map<string, vector<string>>::const_iterator itr;
	for (itr = adjWords.begin(); itr != adjWords.end(); ++itr)
	{
		const pair<string, vector<string>> &entry = *itr;   //pair是一個模板，就是理解成一個對就好了，這裏就是這個pair有兩個元素，first類型是string,second類型是vector<string>
		const vector<string> words = entry.second;
		if (words.size() >= minWords)
		{
			cout << entry.first << "(" << words.size() << "):";
			for (int i = 0; i < words.size(); i++)
			{
				cout << "" << words[i];
			}
			cout << endl;
		}
	}
}

//構造map的一個暴力方法，寫一個函數，比較兩個單詞，是不是隻有一個字母不一樣，之後二層for循環所有的單詞，挨個比較，思路比較簡單，時間複雜度很心酸
bool oneCharoff(const string & word1, const string & word2)
{
	if (word1.length() != word2.length())
	{
		return false;
	}
	
	int diff = 0;
	for (int i = 0; i < word1.length(); i++)
	{
		if (word1[i] != word2[i])
			if (++diff > 1)
				return false;
	}
	
	return diff == 1;
}

//把words形成map
map<string, vector<string>> computeAdjacentWords(const vector<string> & words)
{
	map<string, vector<string>> adjWords;
	for (int i = 0; i < words.size(); i++)
	{
		for (int j = i + 1; j < words.size(); j++)
		{
			if (oneCharoff(words[i], words[j]))
			{
				adjWords[words[i]].push_back(words[j]);
				adjWords[words[j]].push_back(words[i]);
			}
		}
	}
	return adjWords;
}


//第二種，還是一個map，但是鍵變成了單詞的長度，然後把相同長度的放在一起，然後跟上面一樣的過程，加一個循環每個單詞組

map<string, vector<string>> computeAdjacentWords(const vector<string> & words)
{
	map<int, vector<string>> KeyLength;
	map<string, vector<string>> adjWords;
	for (int i = 0; i < words.size(); i++)
	{
		KeyLength[words[i].length()].push_back(words[i]);
	}
	map<int, vector<string>>::const_iterator itr;
	for (itr = KeyLength.begin(); itr != KeyLength.end(); itr++)
	{
		const vector<string> & groupsWords = itr->second;          //這個剛開始沒寫，這樣會方便點，否則容易看暈......
		for (int i = 0; i < groupsWords.size(); i++)
			for (int j = i + 1; j < groupsWords.size(); j++)
			{
			if (oneCharoff((groupsWords)[i], (groupsWords)[j]))
			{
				adjWords[(groupsWords)[i]].push_back((groupsWords)[j]);
				adjWords[(groupsWords)[j]].push_back((groupsWords)[i]);
			}
			}
	}
	return adjWords;
}




//這一種會更快，原理是這樣的，把每個長度相同的詞的集合，按照去掉一個字母一樣分（加一個map）,比如，wine,wife就分到wie組裏，這樣就能很快的構造出最後的那個map（adjWords）
map<string, vector<string>> computeAdjacentWords(const vector<string> & words)
{
	map<int, vector<string>> KeyLength;
	map<string, vector<string>> adjWords;
	for (int i = 0; i < words.size(); i++)
	{
		KeyLength[words[i].length()].push_back(words[i]);
	}
	map<int, vector<string>>::const_iterator itr;
	for (itr = KeyLength.begin(); itr != KeyLength.end(); itr++)
	{
		const vector<string> & groupsWords = itr->second;          
		int groupLen = itr->first;
		for (int i = 0; i < groupLen; i++)
		{
			map<string, vector<string>> Same;
			for (int j = 0; j < groupsWords.size(); j++)
			{
				string rep = groupsWords[j];
				rep.erase(i, 1);
				Same[rep].push_back(groupsWords[j]);
			}
			map<string, vector<string>>::const_iterator itr2;
			for (itr2 = Same.begin(); itr2 != Same.end(); itr2++)
			{
				const vector<string>& sepreate = itr2->second;
				if (sepreate.size() >= 2)
					for (int p = 0; p < sepreate.size(); p++)
						for (int q = p + 1; q < sepreate.size(); q++)
						{
							adjWords[sepreate[p]].push_back(sepreate[q]);
							adjWords[sepreate[q]].push_back(sepreate[p]);
						}
			}
		}
	}
	return adjWords;
}