Thursday, November 27, 2014

[LeetCode] Word Ladder I, II

Word Ladder I

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.


Word Ladder II

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) 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"]
Return
  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]
Note:
  • All words have the same length.
  • All words contain only lowercase alphabetic characters.


思路:

LeetCode中为数不多的考图的难题。尽管题目看上去像字符串匹配题,但从“shortest transformation sequence from start to end”还是能透露出一点图论中最短路径题的味道。如何转化?

1. 将每个单词看成图的一个节点。
2. 当单词s1改变一个字符可以变成存在于字典的单词s2时,则s1与s2之间有连接。
3. 给定s1和s2,问题I转化成了求在图中从s1->s2的最短路径长度。而问题II转化为了求所有s1->s2的最短路径。

无论是求最短路径长度还是求所有最短路径,都是用BFS。在BFS中有三个关键步骤需要实现:

1. 如何找到与当前节点相邻的所有节点。
这里可以有两个策略:
(1) 遍历整个字典,将其中每个单词与当前单词比较,判断是否只差一个字符。复杂度为:n*w,n为字典中的单词数量,w为单词长度。
(2) 遍历当前单词的每个字符x,将其改变成a~z中除x外的任意一个,形成一个新的单词,在字典中判断是否存在。复杂度为:26*w,w为单词长度。
这里可以和面试官讨论两种策略的取舍。对于通常的英语单词来说,长度大多小于100,而字典中的单词数则往往是成千上万,所以策略2相对较优。

2. 如何标记一个节点已经被访问过,以避免重复访问。
可以将访问过的单词从字典中删除。

3. 一旦BFS找到目标单词,如何backtracking找回路径?



Word Ladder I

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class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        dict.insert(end);
        queue<pair<string,int>> q;
        q.push(make_pair(start,1));
        while(!q.empty()) {
            string s = q.front().first;
            int len = q.front().second;
            if(s==end) return len;
            q.pop();
            vector<string> neighbors = findNeighbors(s, dict);
            for(int i=0; i<neighbors.size(); i++) 
                q.push(make_pair(neighbors[i],len+1));
        }
        return 0;
    }
    
    vector<string> findNeighbors(string s, unordered_set<string> &dict) {
        vector<string> ret;
        for(int i=0; i<s.size(); i++) {
            char c = s[i];
            for(int j=0; j<26; j++) {
                if(c=='a'+j) continue;
                s[i] = 'a'+j;
                if(dict.count(s)) {
                    ret.push_back(s);    
                    dict.erase(s);    
                }
            }
            s[i] = c;
        }
        return ret;
    }
};


Word Ladder II


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