[LeetCode] Word Ladder

[Problem]

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.

[Analysis]

深度优先搜索

[Solution]

class Solution {
public:
int ladderLength(string start, string end, unordered_set<string> &dict) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(start == end){
return 1;
}

// initial
int cnt = 1, child = 0, res = 0;
queue<string> myQueue;
set<string> used;
myQueue.push(start);
used.insert(start);
if(dict.find(end) == dict.end())
dict.insert(end);

// BFS
while(!myQueue.empty()){
string word = myQueue.front();
myQueue.pop();
cnt--;
if(cnt == 0){
res++;
}

// match
if(word == end){
if(cnt > 0)res++;
return res;
}

// DFS
for(int i = 0; i < word.length(); ++i){
string tmp = word;
for(int j = 0; j < 26; ++j){
tmp[i] = char('a' + j);
if(dict.find(tmp) != dict.end() && used.find(tmp) == used.end()){
myQueue.push(tmp);
used.insert(tmp);
child++;
}
}
}

// next level
if(cnt == 0){
cnt = child;
child = 0;
}
}
return 0;
}
};

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