Merge pull request #2777 from Jasonyou-boy/feature/nextString

feat(图论): 字符串接龙 Java优化版 并修复 beginStr==endStr 边界条件

Author: youngyangyang04

problems/kamacoder/0098.所有可达路径.md

@@ -277,7 +277,7 @@ ACM格式大家在输出结果的时候，要关注看看格式问题，特别
 
 有录友可能会想，ACM格式就是麻烦，有空格没有空格有什么影响，结果对了不就行了？ 
 
-ACM模式相对于核心代码模式（力扣） 更考验大家对代码的掌控能力。 例如工程代码里，输出输出都是要自己控制的。这也是为什么大公司笔试，都是ACM模式。 
+ACM模式相对于核心代码模式（力扣） 更考验大家对代码的掌控能力。 例如工程代码里，输入输出都是要自己控制的。这也是为什么大公司笔试，都是ACM模式。 
 
 以上代码中，结果都存在了 result数组里（二维数组，每一行是一个结果），最后将其打印出来。（重点看注释） 
 

problems/kamacoder/0100.岛屿的最大面积.md

@@ -222,121 +222,127 @@ public:
 
 ## 其他语言版本
 
-### Java 
-DFS
+### Java
+
 ```java
-//这里的实现为主函数处理每个岛屿的第一块陆地 方式
-//所以是主函数直接置count为1，剩余的交给dfs来做。
 import java.util.*;
+import java.math.*;
+
+/**
+ * DFS版
+ */
 public class Main{
-    static int[][] dir = {{0,-1}, {1,0}, {0,1}, {-1, 0}};//四个方向
-    static int count = 0;
-    public static void dfs(boolean[][] visited, int x, int y, int[][] grid){
-        for(int i = 0; i < 4; i++){
-            int nextX = x + dir[i][0];
-            int nextY = y + dir[i][1];
-            if(nextX < 0 || nextY < 0 || nextY >= grid[0].length || nextX >= grid.length){
-                continue;
-            }
-            if(!visited[nextX][nextY] && grid[nextX][nextY] == 1){
-                count++;
-                visited[nextX][nextY] = true;
-                dfs(visited, nextX, nextY, grid);
-            }
-        }
-    }
+
+    static final int[][] dir={{0,1},{1,0},{0,-1},{-1,0}};
+    static int result=0;
+    static int count=0;
+
     public static void main(String[] args){
-        Scanner in = new Scanner(System.in);
-        int n = in.nextInt();
-        int m = in.nextInt();
-        int[][] grid = new int[n][m];
-        for(int i = 0; i < n; i++){
-            for(int j = 0; j < m; j++){
-                grid[i][j] = in.nextInt();
+        Scanner scanner = new Scanner(System.in);
+        int n = scanner.nextInt();
+        int m = scanner.nextInt();
+        int[][] map = new int[n][m];
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                map[i][j]=scanner.nextInt();
             }
         }
-        
-        int result = 0;
         boolean[][] visited = new boolean[n][m];
-        for(int i = 0; i < n; i++){
-            for(int j = 0; j < m; j++){
-                if(!visited[i][j] && grid[i][j] == 1){
-                    visited[i][j] = true;
-                    count = 1;
-                    dfs(visited, i, j, grid);
-                    //dfs遍历完了一座岛屿，就比较count和result，保留最大的
-                    result = Math.max(result, count);                                         
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                if(!visited[i][j]&&map[i][j]==1){
+                    count=0;
+                    dfs(map,visited,i,j);
+                    result= Math.max(count, result);
                 }
             }
         }
         System.out.println(result);
     }
+
+    static void dfs(int[][] map,boolean[][] visited,int x,int y){
+                count++;
+                visited[x][y]=true;
+                for (int i = 0; i < 4; i++) {
+                    int nextX=x+dir[i][0];
+                    int nextY=y+dir[i][1];
+                    //水或者已经访问过的跳过
+                    if(nextX<0||nextY<0
+                    ||nextX>=map.length||nextY>=map[0].length
+                    ||visited[nextX][nextY]||map[nextX][nextY]==0)continue;
+                    
+                    dfs(map,visited,nextX,nextY);
+                }
+            }
 }
 ```
-BFS
+
 ```java
 import java.util.*;
-public class Main{
-    static int[][] dir = {{0,-1}, {1,0}, {0,1}, {-1, 0}};//下右上左的顺序
-    static int count = 0;
-    public static void bfs(boolean[][] visited, int x, int y, int[][] grid){
-        Queue<pair> queue = new LinkedList<pair>();
-        queue.add(new pair(x,y));
-        count = 1; //该岛屿的第一块陆地被visit了
-        
-        //对这个岛屿的所有都入队，除非上下左右都没有未访问的陆地
-        while(!queue.isEmpty()){
-            int curX = queue.peek().x;
-            int curY = queue.poll().y;
-            //对每块陆地都进行上下左右的入队和计算（遍历），自然就是按广度优先了
-            for(int i = 0; i < 4; i++){
-                int nextX = curX + dir[i][0];
-                int nextY = curY + dir[i][1];
-                if(nextX < 0 || nextY < 0 || nextX >= grid.length || nextY >= grid[0].length){
-                    continue;
-                }
-                if(!visited[nextX][nextY] && grid[nextX][nextY] == 1){
-                    count++;
-                    queue.add(new pair(nextX, nextY));
-                    visited[nextX][nextY] = true;
-                }
-            }
-        }
-    }
-    
-    static class pair{
+import java.math.*;
+
+/**
+ * BFS版
+ */
+public class Main {
+    static class Node {
         int x;
         int y;
-        pair(int x, int y){
+
+        public Node(int x, int y) {
             this.x = x;
             this.y = y;
         }
     }
-    
-    public static void main(String[] args){
-        Scanner in = new Scanner(System.in);
-        int n = in.nextInt();
-        int m = in.nextInt();
-        int[][] grid = new int[n][m];
-        for(int i = 0; i < n; i++){
-            for(int j = 0; j < m; j++){
-                grid[i][j] = in.nextInt();
+
+    static final int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
+    static int result = 0;
+    static int count = 0;
+
+    public static void main(String[] args) {
+        Scanner scanner = new Scanner(System.in);
+        int n = scanner.nextInt();
+        int m = scanner.nextInt();
+        int[][] map = new int[n][m];
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                map[i][j] = scanner.nextInt();
             }
         }
-        int result = 0;
         boolean[][] visited = new boolean[n][m];
-        for(int i = 0; i < n; i++){
-            for(int j = 0; j < m; j++){
-                if(!visited[i][j] && grid[i][j] == 1){
-                    visited[i][j] = true;
-                    bfs(visited, i, j, grid);
-                    result = Math.max(result, count);
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                if (!visited[i][j] && map[i][j] == 1) {
+                    count = 0;
+                    bfs(map, visited, i, j);
+                    result = Math.max(count, result);
+
                 }
             }
         }
         System.out.println(result);
     }
+
+    static void bfs(int[][] map, boolean[][] visited, int x, int y) {
+        Queue<Node> q = new LinkedList<>();
+        q.add(new Node(x, y));
+        visited[x][y] = true;
+        count++;
+        while (!q.isEmpty()) {
+            Node node = q.remove();
+            for (int i = 0; i < 4; i++) {
+                int nextX = node.x + dir[i][0];
+                int nextY = node.y + dir[i][1];
+                if (nextX < 0 || nextY < 0 || nextX >= map.length || nextY >= map[0].length || visited[nextX][nextY] || map[nextX][nextY] == 0)
+                    continue;
+                q.add(new Node(nextX, nextY));
+                visited[nextX][nextY] = true;
+                count++;
+            }
+        }
+    }
 }
+
 ```
 ### Python
 

problems/kamacoder/0110.字符串接龙.md

@@ -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);
-
-            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);
+    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);
+    }
+
+    /**
+     * 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);
-    }
 }
 
 ```