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[Leetcode]127. Word Ladder(C++)

题目描述

题目链接:127. Word Ladder

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

  • Every adjacent pair of words differs by a single letter.
  • Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
  • sk == endWord
    Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 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, and wordList[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:

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#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