LeetCode-Word Ladder

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.

将每差一个字母的两个字符串连起来可以构成图，找图上给定起点终点的最短距离，用BFS

注意unordered_set，map, queue的使用

class Solution {
public:
	int ladderLength(string start, string end,  std::unordered_set<string> &dict) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		queue<string> q;
		map<string,int> m;
		q.push(start);
		m[start]=0;
		while(q.size()!=0){
			string top=q.front();
			q.pop();
			string s=top;
			for(int i=0;i<top.length();i++){
				char c=top[i];
				for(int j='a';j<c;j++){
				    s[i]=j;
					if(s==end)return m[top]+2;
					if(dict.find(s)!=dict.end()&&m.find(s)==m.end()){
						m[s]=m[top]+1;
						q.push(s);
					}
				}
				for(int j=c+1;j<'z';j++){
				    s[i]=j;
					if(s==end)return m[top]+2;
					if(dict.find(s)!=dict.end()&&m.find(s)==m.end()){
						m[s]=m[top]+1;
						q.push(s);
					}
				}
				s[i]=c;
			}
		}
		return 0;
	}
};