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112 lines (103 loc) · 3.55 KB
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// Source : https://oj.leetcode.com/problems/word-ladder/
// Author : Hao Chen
// Date : 2014-10-12
/**********************************************************************************
*
* 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.
*
*
**********************************************************************************/
// ---------------------------
// BFS non-recursive method
// ---------------------------
//
// Using BFS instead of DFS is becasue the solution need the shortest transformation path.
//
// So, we can change every char in the word one by one, until find all possible transformation.
//
// Keep this iteration, we will find the shorest path.
//
// For example:
//
// start = "hit"
// end = "cog"
// dict = ["hot","dot","dog","lot","log","dit","hig", "dig"]
//
// +-----+
// +-------------+ hit +--------------+
// | +--+--+ |
// | | |
// +--v--+ +--v--+ +--v--+
// | dit | +-----+ hot +---+ | hig |
// +--+--+ | +-----+ | +--+--+
// | | | |
// | +--v--+ +--v--+ +--v--+
// +----> dot | | lot | | dig |
// +--+--+ +--+--+ +--+--+
// | | |
// +--v--+ +--v--+ |
// +----> dog | | log | |
// | +--+--+ +--+--+ |
// | | | |
// | | +--v--+ | |
// | +--->| cog |<-- + |
// | +-----+ |
// | |
// | |
// +----------------------------------+
//
// 1) queue <== "hit"
// 2) queue <== "dit", "hot", "hig"
// 3) queue <== "dot", "lot", "dig"
// 4) queue <== "dog", "log"
//
class Solution {
public:
int ladderLength(string start, string end, unordered_set<string> &dict) {
// Using a map for two purposes:
// 1) store the distince so far.
// 2) remove the duplication
map<string, int> dis;
dis[start] = 1;
queue<string> q;
q.push(start);
while(!q.empty()){
string word = q.front();
q.pop();
if (word == end) {
break;
}
for (int i=0; i<word.size(); i++){
string temp = word;
for(char c='a'; c<='z'; c++){
temp[i] = c;
if (dict.count(temp)>0 && dis.count(temp)==0){
dis[temp] = dis[word] + 1;
q.push(temp);
}
}
}
}
return (dis.count(end)==0) ? 0 : dis[end];
}
};