feate: 字符串接龙 Java优化版并修复存在问题

1.使用null标志分层
2.寻找邻接节点时,回滚修改,避免多次数组复制
3.修复边界条件:beginStr==endStr时,应该输出0;
This commit is contained in:
Jasonyou
2024-10-06 03:04:09 +08:00
parent 4eef0863a2
commit 9c7131e253

View File

@ -152,66 +152,70 @@ int main() {
## 其他语言版本
### Java
### Java
```Java
import java.util.*;
public class Main {
// BFS方法
public static int ladderLength(String beginWord, String endWord, List<String> wordList) {
// 使用set作为查询容器效率更高
HashSet<String> set = new HashSet<>(wordList);
// 声明一个queue存储每次变更一个字符得到的且存在于容器中的新字符串
Queue<String> queue = new LinkedList<>();
// 声明一个hashMap存储遍历到的字符串以及所走过的路径path
HashMap<String, Integer> visitMap = new HashMap<>();
queue.offer(beginWord);
visitMap.put(beginWord, 1);
while (!queue.isEmpty()) {
String curWord = queue.poll();
int path = visitMap.get(curWord);
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
scanner.nextLine();
String beginStr = scanner.next();
String endStr = scanner.next();
scanner.nextLine();
List<String> wordList = new ArrayList<>();
wordList.add(beginStr);
wordList.add(endStr);
for (int i = 0; i < n; i++) {
wordList.add(scanner.nextLine());
}
int count = bfs(beginStr, endStr, wordList);
System.out.println(count);
}
for (int i = 0; i < curWord.length(); i++) {
char[] ch = curWord.toCharArray();
// 每个位置尝试26个字母
for (char k = 'a'; k <= 'z'; k++) {
ch[i] = k;
String newWord = new String(ch);
if (newWord.equals(endWord)) return path + 1;
// 如果这个新字符串存在于容器且之前未被访问到
if (set.contains(newWord) && !visitMap.containsKey(newWord)) {
visitMap.put(newWord, path + 1);
queue.offer(newWord);
/**
* 广度优先搜索-寻找最短路径
*/
public static int bfs(String beginStr, String endStr, List<String> wordList) {
int len = 1;
Set<String> set = new HashSet<>(wordList);
Set<String> visited = new HashSet<>();
Queue<String> q = new LinkedList<>();
visited.add(beginStr);
q.add(beginStr);
q.add(null);
while (!q.isEmpty()) {
String node = q.remove();
//上一层结束,若下一层还有节点进入下一层
if (node == null) {
if (!q.isEmpty()) {
len++;
q.add(null);
}
continue;
}
char[] charArray = node.toCharArray();
//寻找邻接节点
for (int i = 0; i < charArray.length; i++) {
//记录旧值,用于回滚修改
char old = charArray[i];
for (char j = 'a'; j <= 'z'; j++) {
charArray[i] = j;
String newWord = new String(charArray);
if (set.contains(newWord) && !visited.contains(newWord)) {
q.add(newWord);
visited.add(newWord);
//找到结尾
if (newWord.equals(endStr)) return len + 1;
}
}
charArray[i] = old;
}
}
return 0;
}
public static void main (String[] args) {
/* code */
// 接收输入
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
sc.nextLine();
String[] strs = sc.nextLine().split(" ");
List<String> wordList = new ArrayList<>();
for (int i = 0; i < N; i++) {
wordList.add(sc.nextLine());
}
// wordList.add(strs[1]);
// 打印结果
int result = ladderLength(strs[0], strs[1], wordList);
System.out.println(result);
}
}
```