【leetcode】Word Ladder

Word Ladder

鏈接：https://oj.leetcode.com/problems/word-ladder/

 

描述：

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

方法：

BFS

第一個搜索到的結果的層數+2即爲結果。

由於都是小寫字符，針對每一個字符串的每一位依次從 ‘a'到’z'替換原來的字符，如果替換後的字符串爲end串，則結束，如果不是查看字典中是否有，如果有放在隊列中。

實現代碼一：

該代碼沒有標記字典中訪問過的字符串，導致內存超了。

int ladderLength(string start, string end, set<string> &dict)
{
	if(start.size()  != end.size()) return 0;
	if(start.empty() || end.empty()) return 1;
	if(dict.empty()) return 0;
	int len = start.length();
	if( start.compare(end) == 0) return 2;
	queue<string, list<string>> q;
	q.push(start);
	q.push("#");
	int result = 0;

	while(!q.empty())
	{
		string s= q.front();
		q.pop();
		if( s.compare("#") == 0){
			result++;
			if( !q.empty())
				q.push("#");
			continue;
		}
		for(int i=0; i < len; ++i)
		{
			char oc = s[i];
			for(int j = 0; j < 26; ++j)
			{
				s[i] = 'a' + j;
				if( s[i] == oc)
					continue;
				if( s.compare(end) == 0)
					return result+2;
				else if(dict.count(s) > 0)
					q.push(s);
			}
			s[i] = oc;
		}
	}

	return 0;
}

實現二：

加了訪問標誌。Accept。

代碼如下：

int ladderLength1(string start, string end, set<string> &dict)
{
	if(start.size()  != end.size()) return 0;
	if(start.empty() || end.empty()) return 1;
	if(dict.empty()) return 0;
	int len = start.length();
	if( start.compare(end) == 0) return 2;
	queue<string> q;
	map<string, bool> flag;
	q.push(start);
	q.push("#");
	int result = 0;

	while(!q.empty())
	{
		string s= q.front();
		q.pop();
		if( s.compare("#") == 0){
			result++;
			if( !q.empty())
				q.push("#");
			continue;
		}
		for(int i=0; i < len; ++i)
		{
			char oc = s[i];
			for(int j = 0; j < 26; ++j)
			{
				s[i] = 'a' + j;
				if( s[i] == oc)
					continue;
				if( s.compare(end) == 0)
					return result+2;
				else if(dict.count(s) > 0 && flag.count(s) <= 0)
				{
					flag[s] = true;
					q.push(s);
				}
			}
			s[i] = oc;
		}
	}

	return 0;
}

實現三：

上面代碼使用 “#” 來表示一層結束，也就是表示層數。

下面代碼使用兩個棧交替，每一個棧代表一層的節點情況。

代碼如下：

int ladderLength2(string start, string end, set<string> &dict) {
	if(start.size()  != end.size()) return 0;
	if(start.empty() || end.empty()) return 1;
	if(dict.empty()) return 0;
	queue<string> current,next;
	map<string,bool> flag;
	int result = 1;
	bool found = false;
	current.push(start);
	while(!current.empty()){
		while(!current.empty()){
			string origin = current.front();
			current.pop();
			int len = origin.length();
			for(int i=0; i< len; i++){
				for(char c='a'; c <= 'z';++c){
					swap(c,origin[i]);
					if(origin== end){
						return result+1;
					}
					if(dict.count(origin) > 0 && flag.find(origin) == flag.end()){
						next.push(origin);
						flag[origin] = true;
					}
					swap(c,origin[i]);
				}
			}
		}
		swap(current,next);
		result++;
	}
	return 0;
}