LeetCode: Word Ladder

最新推荐文章于 2017-12-31 22:21:58 发布

wyc1230

最新推荐文章于 2017-12-31 22:21:58 发布

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本文介绍了一种算法，用于寻找两个单词之间的最短转换路径。给定起始词、目标词及词典，算法通过逐个替换字母的方式找到从起始词到目标词的最短路径。文中提供了两种实现方案：一种采用单向广度优先搜索，另一种采用双向广度优先搜索提高效率。

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

Only one letter can be changed at a time
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.

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        queue<string> nodeQ;
        queue<int> depthQ;
        nodeQ.push(start);
        depthQ.push(0);
        while(!nodeQ.empty())
        {
            string cur = nodeQ.front();
            nodeQ.pop();
            int depth = depthQ.front();
            depthQ.pop();
            for(int i = 0; i < cur.size(); i++)
            {
                string tmp = cur;
                for(char a = 'a'; a < 'z'; a++)
                {
                    tmp[i] = a;
                    if(cur == end)
                        return depth + 1;
                    else if(dict.find(tmp) != dict.end())
                    {
                        nodeQ.push(tmp);
                        depthQ.push(depth + 1);
                        dict.erase(tmp);
                    }
                }
            }
        }
        return 0;
    }
};

Round 2:

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        queue<string> startSet[2];
        int result = 2;
		int flip = 0;
        startSet[flip].push(start);
        while(!startSet[flip].empty())
        {
            string temp = startSet[flip].front();
            startSet[flip].pop();
            if(findInDic(temp, dict, startSet[1-flip], end))
                return result;
            if(startSet[flip].empty())
            {
			    flip = 1-flip;
                result++;
            }
        }
        return 0;
    }
private:
    bool findInDic(string target, unordered_set<string> &dict, queue<string> &result, string end)
    {
        for(int i = 0; i < target.size(); i++)
        {
            for(char j = 'a'; j <= 'z'; j++)
            {
                string temp = target;
                temp[i] = j;
                if(temp == end)
                    return true;
                if(dict.find(temp) != dict.end())
                {
                    result.push(temp);
                    dict.erase(temp);
                }
            }
        }
        return false;
    }
};