LeetCode 127 单词接龙_leetcode 单词接龙 c语言

单词接龙

时间超时

class Solution {
public:
    bool isvalue(string s1, string s2)
{
	int len = s1.size();
	int count = 0;
	for (int i = 0; i<len; ++i)
	{
		if (s1[i] != s2[i])
			++count;
	}
	if (count == 1)
		return true;
	return false;

}

int counta = 1;
int BFS(string beginWord, string endWord, vector<string>& wordList, map<string, vector<string> > neigh, map<string, int>& flag)
{
	deque<pair<string,int> > S;
	S.push_back(pair<string, int>(beginWord,1) );//当前节点进入队列中        
	while (!S.empty())
	{
		string temp = S.front().first;//当前遍历的节点
		int step = S.front().second;
		vector<string> svec = neigh[temp];
		S.pop_front();
		if (temp == endWord)
		{
			return step;
		}
		for (int i = 0; i<svec.size(); ++i)
		{
			string s1 = svec[i];
			if (flag[s1] == 0)
			{
				//++counta;
								
				flag[s1] = 1;
				S.push_back(pair<string,int>(s1,step+1));
			}
		}
	}
	return 0;
	
}

int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
	map<string, vector<string> > neigh;
	map<string, int> flag;//标志
	int len = wordList.size();
	vector<string> v_sa;
	neigh[beginWord] = v_sa;
	for (int i = 0; i < len; ++i)
	{
		if (isvalue(beginWord, wordList[i]))
			neigh[beginWord].push_back(wordList[i]);
	}
	for (int i = 0; i < len; ++i)
	{
		vector<string> v_s;
		string temp1 = wordList[i];
		neigh[temp1] = v_s;
	}
	for (int i = 0; i<len; ++i)
	{
		string temp1 = wordList[i];		
		for (int j =i+1; j<len; ++j)
		{
			if (isvalue(temp1, wordList[j]))
			{
				neigh[temp1].push_back(wordList[j]);
				neigh[wordList[j]].push_back(temp1);
			}
		}
	}
	for (int i = 0; i < len; ++i)
		flag[wordList[i]] = 0;
	auto res=BFS(beginWord, endWord, wordList, neigh, flag);
	return res;
}
};
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使用另一种方法也是广度优先遍历

class Solution {
public:
       int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
	//使用一定的set和排序
	unordered_set<string> wordlist(wordList.begin(), wordList.end());
	if (wordlist.count(endWord) == 0)
		return 0;
	deque<string> S;
	S.push_back(beginWord);
	int l = beginWord.size();
	int step = 0;
	while (!S.empty())
	{
		++step;
		int sslen = S.size();
		
		for (int i = 0; i<sslen; ++i)
		{
			string temp = S.front();
			S.pop_front();
			if (temp == endWord)
				return step;
			for (int i = 0; i<l; ++i)
			{
				string bianhua = temp;
				for (char a = 'a'; a<'z'; ++a)
				{
					bianhua[i] = a;
					if (wordlist.count(bianhua) != 0)
					{
						S.push_back(bianhua);
						wordlist.erase(bianhua);
					}
				}
			}
		}

	}

	return 0;
}
};
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