给定两个单词(beginWord 和 endWord)和一个字典,找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则:
每次转换只能改变一个字母。 转换过程中的中间单词必须是字典中的单词。 说明:
如果不存在这样的转换序列,返回 0。 所有单词具有相同的长度。 所有单词只由小写字母组成。 字典中不存在重复的单词。 你可以假设 beginWord 和 endWord 是非空的,且二者不相同。 示例 1:
输入: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
输出: 5
解释: 一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。 示例 2:
输入: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log"]
输出: 0
解释: endWord "cog" 不在字典中,所以无法进行转换。 通过次数72,245提交次数162,120
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/word-ladder 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
所以这个博文 还是 分析官方题解 给的答案
class Solution {
public:
unordered_map<string, int> wordId;
vector<vector<int>> edge;
int nodeNum = 0;
void addWord(string& word) {
if (!wordId.count(word)) {
wordId[word] = nodeNum++;
edge.emplace_back();
}
}
void addEdge(string& word) {
addWord(word);
int id1 = wordId[word];
for (char& it : word) {
char tmp = it;
it = '*';
addWord(word);
int id2 = wordId[word];
edge[id1].push_back(id2);
edge[id2].push_back(id1);
it = tmp;
}
}
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
for (string& word : wordList) {
addEdge(word);
}
addEdge(beginWord);
if (!wordId.count(endWord)) {
return 0;
}
vector<int> dis(nodeNum, INT_MAX);
int beginId = wordId[beginWord], endId = wordId[endWord];
dis[beginId] = 0;
queue<int> que;
que.push(beginId);
while (!que.empty()) {
int x = que.front();
que.pop();
if (x == endId) {
return dis[endId] / 2 + 1;
}
for (int& it : edge[x]) {
if (dis[it] == INT_MAX) {
dis[it] = dis[x] + 1;
que.push(it);
}
}
}
return 0;
}
};
作者:LeetCode-Solution
链接:https://leetcode-cn.com/problems/word-ladder/solution/dan-ci-jie-long-by-leetcode-solution/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
class Solution { public: unordered_map<string, int> wordId; vector<vector<int>> edge; int nodeNum = 0;
void addWord(string& word) { if (!wordId.count(word)) { wordId[word] = nodeNum++; edge.emplace_back(); } }
void addEdge(string& word) { addWord(word); int id1 = wordId[word]; for (char& it : word) { char tmp = it; it = '*'; addWord(word); int id2 = wordId[word]; edge[id1].push_back(id2); edge[id2].push_back(id1); it = tmp; } }
for (char& it : word) { char tmp = it; it = '*'; addWord(word); 。。。 it = tmp; }
int id2 = wordId[word]; edge[id1].push_back(id2); edge[id2].push_back(id1);
int ladderLength(string beginWord, string endWord, vector<string>& wordList) { for (string& word : wordList) { addEdge(word); } addEdge(beginWord);
if (!wordId.count(endWord)) { return 0; }
vector<int> dis(nodeNum, INT_MAX); int beginId = wordId[beginWord], endId = wordId[endWord];
dis[beginId] = 0;
queue<int> que; que.push(beginId);
while (!que.empty()) { int x = que.front(); que.pop(); if (x == endId) { return dis[endId] / 2 + 1; }
for (int& it : edge[x]) { if (dis[it] == INT_MAX) { dis[it] = dis[x] + 1; que.push(it); } }
}
return 0; } };
更快 更高 更强
class Solution {
public:
unordered_map<string, int> wordId;
vector<vector<int>> edge;
int nodeNum = 0;
void addWord(string& word) {
if (!wordId.count(word)) {
wordId[word] = nodeNum++;
edge.emplace_back();
}
}
void addEdge(string& word) {
addWord(word);
int id1 = wordId[word];
for (char& it : word) {
char tmp = it;
it = '*';
addWord(word);
int id2 = wordId[word];
edge[id1].push_back(id2);
edge[id2].push_back(id1);
it = tmp;
}
}
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
for (string& word : wordList) {
addEdge(word);
}
addEdge(beginWord);
if (!wordId.count(endWord)) {
return 0;
}
vector<int> disBegin(nodeNum, INT_MAX);
int beginId = wordId[beginWord];
disBegin[beginId] = 0;
queue<int> queBegin;
queBegin.push(beginId);
vector<int> disEnd(nodeNum, INT_MAX);
int endId = wordId[endWord];
disEnd[endId] = 0;
queue<int> queEnd;
queEnd.push(endId);
while (!queBegin.empty() && !queEnd.empty()) {
int queBeginSize = queBegin.size();
for (int i = 0; i < queBeginSize; ++i) {
int nodeBegin = queBegin.front();
queBegin.pop();
if (disEnd[nodeBegin] != INT_MAX) {
return (disBegin[nodeBegin] + disEnd[nodeBegin]) / 2 + 1;
}
for (int& it : edge[nodeBegin]) {
if (disBegin[it] == INT_MAX) {
disBegin[it] = disBegin[nodeBegin] + 1;
queBegin.push(it);
}
}
}
int queEndSize = queEnd.size();
for (int i = 0; i < queEndSize; ++i) {
int nodeEnd = queEnd.front();
queEnd.pop();
if (disBegin[nodeEnd] != INT_MAX) {
return (disBegin[nodeEnd] + disEnd[nodeEnd]) / 2 + 1;
}
for (int& it : edge[nodeEnd]) {
if (disEnd[it] == INT_MAX) {
disEnd[it] = disEnd[nodeEnd] + 1;
queEnd.push(it);
}
}
}
}
return 0;
}
};
作者:LeetCode-Solution
链接:https://leetcode-cn.com/problems/word-ladder/solution/dan-ci-jie-long-by-leetcode-solution/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
int ladderLength(string beginWord, string endWord, vector<string>& wordList) { for (string& word : wordList) { addEdge(word); } addEdge(beginWord); if (!wordId.count(endWord)) { return 0; }
vector<int> disBegin(nodeNum, INT_MAX); int beginId = wordId[beginWord]; disBegin[beginId] = 0; queue<int> queBegin; queBegin.push(beginId); vector<int> disEnd(nodeNum, INT_MAX); int endId = wordId[endWord]; disEnd[endId] = 0; queue<int> queEnd; queEnd.push(endId);
while (!queBegin.empty() && !queEnd.empty()) { int queBeginSize = queBegin.size(); for (int i = 0; i < queBeginSize; ++i) { int nodeBegin = queBegin.front(); queBegin.pop(); if (disEnd[nodeBegin] != INT_MAX) {
return (disBegin[nodeBegin] + disEnd[nodeBegin]) / 2 + 1;
顺向搜 到了 nodeBegin 同时 disEnd[nodeBegin] != INT_MAX 逆向的已经 搜索到 nodeBegin 就返回
} for (int& it : edge[nodeBegin]) { if (disBegin[it] == INT_MAX) { disBegin[it] = disBegin[nodeBegin] + 1; queBegin.push(it);
} } }
int queEndSize = queEnd.size(); for (int i = 0; i < queEndSize; ++i) { int nodeEnd = queEnd.front(); queEnd.pop(); if (disBegin[nodeEnd] != INT_MAX) { return (disBegin[nodeEnd] + disEnd[nodeEnd]) / 2 + 1; } for (int& it : edge[nodeEnd]) { if (disEnd[it] == INT_MAX) { disEnd[it] = disEnd[nodeEnd] + 1; queEnd.push(it); } } } } return 0; }