【Hot100】LeetCode—79. 单词搜索

---

• 原题链接：79. 单词搜索

---

1- 思路

回溯

思路：① 遍历每个单元格（作为起点）、②对每个单元格进行回溯（起点回溯）
1- 起点

• 用两层 for 循环遍历每个单元格，之后对当前单元格进行 dfs

2- 回溯操作

• 由于需要对每个单元格进行回溯，所以 i 和 j的位置，回溯函数必须要知道，因此传参中必须有 i 和 j

---

2- 实现

⭐79. 单词搜索——题解思路

class Solution {
    boolean[][] used ;
    public boolean exist(char[][] board, String word) {
        used = new boolean[board.length][board[0].length];
        int row = board.length;
        int col = board[0].length;
        for(int i = 0 ; i < row;i++){
            for(int j = 0 ; j < col;j++){
                if(backTracing(board,word,0,i,j)){
                    return true;
                }
            }
        }
        return false;
    }

    int[][] dir = {{-1,0},{0,-1},{1,0},{0,1}};
    public boolean backTracing(char[][] board,String word,int index,int i ,int j ){
        // 结果收集
        if(word.length() == index){
            return true;
        }

        // 结束判断
        if( i < 0 || i>=board.length || j<0 || j>=board[0].length || used[i][j] || board[i][j] != word.charAt(index)){
            return false;
        }

        // 回溯
        // 标记为已使用
        used[i][j] = true;
        for(int[] d:dir){
            int nextX = i + d[0];
            int nextY = j + d[1];
            if(backTracing(board,word,index+1,nextX,nextY)){
                return true;
            }
        }
        used[i][j] = false;
        return false;
    }
}

---

3- ACM 实现

public class exist {



    static boolean[][] used ;
    public static boolean exist(char[][] board, String word) {
        used = new boolean[board.length][board[0].length];
        int row = board.length;
        int col = board[0].length;
        for(int i = 0 ; i < row;i++){
            for(int j = 0 ; j < col;j++){
                if(backTracing(board,word,0,i,j)){
                    return true;
                }
            }
        }
        return false;
    }

    static int[][] dir = {{-1,0},{0,-1},{1,0},{0,1}};
    public static boolean backTracing(char[][] board,String word,int index,int i ,int j ){
        // 结果收集
        if(word.length() == index){
            return true;
        }

        // 结束判断
        if( i < 0 || i>=board.length || j<0 || j>=board[0].length || used[i][j] || board[i][j] != word.charAt(index)){
            return false;
        }

        // 回溯
        // 标记为已使用
        used[i][j] = true;
        for(int[] d:dir){
            int nextX = i + d[0];
            int nextY = j + d[1];
            if(backTracing(board,word,index+1,nextX,nextY)){
                return true;
            }
        }
        used[i][j] = false;
        return false;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String boardInput = scanner.nextLine();
        String word = scanner.nextLine();
        char[][] board = parseBoard(boardInput);
        boolean res = exist(board, word);
        System.out.println(res);
    }

    private static char[][] parseBoard(String input) {
        input = input.substring(1, input.length() - 1); // Remove outer brackets
        String[] rows = input.split("],\\[");
        int rowCount = rows.length;
        int colCount = rows[0].split(",").length;
        char[][] board = new char[rowCount][colCount];
        for (int i = 0; i < rowCount; i++) {
            String[] cols = rows[i].replaceAll("[\\[\\]\"]", "").split(",");
            for (int j = 0; j < colCount; j++) {
                board[i][j] = cols[j].charAt(0);
            }
        }
        return board;
    }
}

---