题目描述
题目链接:127. Word Ladder
A transformation sequence from word
beginWord
to wordendWord
using a dictionarywordList
is a sequence of wordsbeginWord -> s1 -> s2 -> ... -> sk
such that:
- Every adjacent pair of words differs by a single letter.
- Every
si
for1 <= i <= k
is in wordList. Note thatbeginWord
does not need to be in wordList.sk == endWord
Given two words,beginWord
andendWord
, and a dictionarywordList
, return the number of words in the shortest transformation sequence frombeginWord
toendWord
, or0
if no such sequence exists.
例子
例子 1
Input:
beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output:5
Explanation:One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.
例子 2
Input:
beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output:0
Explanation:The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.
Constraints
1 <= beginWord.length <= 10
endWord.length == beginWord.length
1 <= wordList.length <= 5000
wordList[i].length == beginWord.length
beginWord
,endWord
, andwordList[i]
consist of lowercase English letters.beginWord != endWord
- All the words in
wordList
are unique.
解题思路
题目要求在字典中找到一个单词序列,其中相邻单词差异字母数量为 1, 序列中第一个单词和给定起始单词相差 1,最后一个单词为给定结束单词。可以创建一个图,然后用 endWord
作为第一个节点进行广度优先搜索(因为 beginWord
不再序列中),最后找到符合条件的最后一个单词时返回迭代层数即可。
代码如下:
#include <queue>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>
class Solution {
public:
int ladderLength(std::string beginWord, std::string endWord,
std::vector<std::string>& wordList) {
// construct graph
std::unordered_map<int, std::vector<int>> graph;
std::queue<int> candidates;
for (int i = 0; i < wordList.size(); ++i) {
if (wordList[i] == endWord) {
candidates.push(i);
}
graph[i] = {};
for (int j = 0; j < i; ++j) {
if (countDiff(wordList[i], wordList[j]) == 1) {
graph[i].push_back(j);
graph[j].push_back(i);
}
}
}
int step = 1;
std::unordered_set<int> used;
while (!candidates.empty()) {
std::queue<int> next;
while (!candidates.empty()) {
int idx = candidates.front();
candidates.pop();
// cout << wordList[idx] << endl;
if (countDiff(wordList[idx], beginWord) == 1) {
return ++step;
}
used.insert(idx);
for (auto neighbor : graph[idx]) {
if (used.count(neighbor) == 0) {
next.push(neighbor);
}
}
}
step++;
candidates = next;
}
return 0;
}
private:
int countDiff(const std::string& first, const std::string& second) {
if (first.size() != second.size()) {
return -1;
}
int count = 0;
for (int i = 0; i < first.size(); ++i) {
count += (first[i] != second[i]);
}
return count;
}
};
- 时间复杂度:
O(|V| + |E|)
- 空间复杂度:
O(|E|)
GitHub 代码同步地址: 127.WordLadder.cpp
其他题目: GitHub: Leetcode-C++-Solution 博客: Leetcode-Solutions