标签：des style blog 使用 io strong for ar art
Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
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:
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
}
};
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:
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:
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;
}
};
18. Word Ladder && Word Ladder II
标签：des style blog 使用 io strong for ar art
原文地址：http://www.cnblogs.com/liyangguang1988/p/3933973.html
