【DSF】Word Search-优快云博客

本文探讨了如何在一个二维网格中寻找特定单词的存在性，通过深度优先搜索（DFS）算法实现，确保避免重复访问同一字母单元格。

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

解法：DFS，上下左右深搜，注意为了防止循环搜，比如：“aa","aaa"的情况，dfs的时候若字符匹配则需先替换成空字符，dfs之后再回溯

public class Solution {
    private boolean find = false;
    private char[][] board;
    private String word;
    private int row, col;
    
    public void dfs(int x, int y, int pos){
        if(find == true) return;
        if(pos == word.length()){
            find = true;
            return;
        }
        if(x<0 || x>=row || y<0 || y>=col || board[x][y] != word.charAt(pos)) return;
        
        char c = board[x][y];
        board[x][y] = ' ';
        
        dfs(x-1, y, pos+1);
        dfs(x+1, y, pos+1);
        dfs(x, y-1, pos+1);
        dfs(x, y+1, pos+1);
        
        board[x][y] = c;//回溯
    }
    
    public boolean exist(char[][] board, String word) {
        this.board = board;
        this.word = word;
        
        if(board == null || board.length == 0) return false;
        
        row = board.length;
        col = board[0].length;
        
        int pos = 0;
        for(int i=0; i<row&&!find; i++){
            for(int j=0; j<col&&!find; j++){
                dfs(i, j, pos);
            }
        }
        return find;
    }
}