18. Word Ladder && Word Ladder II

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.

思想：宽度优先搜索(BFS)即可。（可从头往尾部搜，也可从尾往头部搜，）。

我的方案中，使用两个 hash_set 分别存储当前层和下一层结点，另一个 hash_set存储之前遍历过的结点。

class Solution {
public:
	int ladderLength(string start, string end, unordered_set<string> &dict) {
		int ans = 0;
		unordered_set<string> previousNodes;
		vector<unordered_set<string> > node_levels(2);
		int curLevel = 0; // which is index belong to vector node_levels.
		node_levels[curLevel].insert(end);
		ans++;
		unordered_set<string>::iterator it;
		while(!node_levels[curLevel].empty()) {
			for(it = node_levels[curLevel].begin(); it != node_levels[curLevel].end(); ++it) {
				for(size_t i = 0; i < it->size(); ++i) {
					string node(*it);
					for(node[i] = 'a'; node[i] <= 'z'; ++node[i]) {
						if(node == start) return (ans+1); // output 1
						if(previousNodes.count(node) || node_levels[curLevel].count(node) || node[i] == (*it)[i] || !dict.count(node))
						    continue;
						node_levels[!curLevel].insert(node);
					}
				}
				previousNodes.insert(*it);
			}
			node_levels[curLevel].clear();
			curLevel = !curLevel;
			ans++;
		}
		return 0; // output 2
	}
};

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.

思想： 在 I 的基础之上， 加入 hash_map 记下每条边， 从 end 开始搜索，建立以 start 为源点，end 为汇点的图，然后从 start 开始深搜即可。

typedef pair<string, string> PAIR;
void  getSolution(string &end, string& word, unordered_multimap<string, string> &map, vector<vector<string> > &vec, vector<string> &vec2) {
	if(word == end) {
		vec.push_back(vec2);
		vec.back().push_back(word);
		return;
	}
	pair<unordered_map<string, string>::iterator, unordered_map<string, string>::iterator> ret;
	ret = map.equal_range(word);
	while(ret.first != ret.second) {
		vec2.push_back(ret.first->first);
		getSolution(end, ret.first->second, map, vec, vec2);
		vec2.pop_back();
		ret.first++;
	}
}
class Solution {
public:
	vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
		vector<vector<string>> vec;
		unordered_multimap<string, string> edges;
		unordered_set<string> previousNodes;
		vector<unordered_set<string> > node_levels(2);
		int curLevel = 0; // an index belong to vector node_levels
		node_levels[curLevel].insert(end);
		unordered_set<string>::iterator it;
		while(!node_levels[curLevel].empty() && node_levels[curLevel].count(start) == 0) {
			for(it = node_levels[curLevel].begin(); it != node_levels[curLevel].end(); ++it) {
				for(size_t i = 0; i < it->size(); ++i) {
					string node(*it);
					for(node[i] = 'a'; node[i] <= 'z'; ++node[i]) {
						if(node == start) {
							node_levels[1-curLevel].insert(node);
							edges.insert(PAIR(start, *it));
							break;
						}
						if(previousNodes.count(node) || node_levels[curLevel].count(node) || dict.count(node) == 0)
						    continue;
						node_levels[1-curLevel].insert(node);
						edges.insert(PAIR(node, *it));
					}
				}
				previousNodes.insert(*it);
			}
			node_levels[curLevel].clear();
			curLevel = !curLevel;
		}
		previousNodes.clear();
		if(node_levels[curLevel].empty()) return vec;
		vector<string> vec2;
		getSolution(end, start, edges, vec, vec2);
		return vec;
	}
};