每日一练_105(2021.9.21) 单词搜索（leetcode）。_每日一练java 单词搜索-优快云博客

本文介绍了一种使用回溯法在二维字符网格中搜索指定单词的方法。通过递归深度优先搜索，遍历网格上每个可能路径，检查是否能构成目标单词。提供了完整的Java实现代码及示例。

唉，每天也就靠看答案解析，来混混日子了。天哪，啥都不会，啥也不是......

推荐作者详解：作者：liweiwei1419
链接：https://leetcode-cn.com/problems/word-search/solution/zai-er-wei-ping-mian-shang-shi-yong-hui-su-fa-pyth/
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。

个人微改：

class search{
   private static final int[][] DIRECTIONS = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
private int rows;
private int cols;
private int len;
private boolean[][] visited;
private char[] charArray;
private char[][] board;
   public boolean exist(char[][] board,String word) {
       this.rows = board.length;
       if(rows==0) return false;
       this.cols = board[0].length;
       this.visited = new boolean[rows][cols];
       this.len = word.length();
       this.charArray = word.toCharArray();
       this.board = board;
       for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (dfs(i, j, 0)) {
return true;
}
}
}
       return false;
   }

   public boolean inArea(int x,int y) {
       return x>=0 && x<rows && y>=0 && y<cols;
   }

   public boolean dfs(int x,int y,int begin) {
       if(begin == len-1) {
           return board[x][y] == charArray[begin];
       }
       if(board[x][y] == charArray[begin]) {
           visited[x][y] = true;
           for(int[] direction:DIRECTIONS) {
               int newX = x + direction[0];
               int newY = y + direction[1];
               if(inArea(newX,newY) && !visited[newX][newY]) {
                   if(dfs(newX,newY,begin+1)) {
                       return true;
                   }
               }
           }
           visited[x][y] = false;
       }
       return false;
   }
}
public class WordSearch {
   public static void main(String args[]) {
       char[][] board1 = new char[][] {{'A','B','C','D'},{'S','F','C','S'},{'A','D','E','E'}};
       String str1 = "ABCCED";
       String str2 = "ABCB";
       search hah = new search();
       System.out.println(hah.exist(board1, str1));
       System.out.println(hah.exist(board1, str2));
   }
}