79. 单词搜索 (opens new window)

1.DFS+回溯

解题思路：每次搜索当前节点(x,y)四个方向的相邻节点，如果该节点的字符等于单词的当前index字符，那么进入该节点继续搜索。

class Solution {
    public boolean exist(char[][] board, String word) {
        boolean[][] visited=new boolean[board.length][board[0].length];
        for(int i=0;i<board.length;i++){
            for(int j=0;j<board[0].length;j++){
                if(board[i][j]==word.charAt(0)){
                    visited[i][j]=true;
                    if(dfs(board,visited,word,1,i,j)) return true;
                    visited[i][j]=false;
                }
            }
        }
        return false;
    }
    public boolean dfs(char[][] board,boolean[][] visited,String word,int index,int x,int y){
        if(index==word.length()) return true;
        if(x>=0&&x<board.length&&y+1>=0&&y+1<board[0].length&&!visited[x][y+1]&&board[x][y+1]==word.charAt(index)){
            visited[x][y+1]=true;
            if(dfs(board,visited,word,index+1,x,y+1)) return true;
            visited[x][y+1]=false;
        }
       if(x+1>=0&&x+1<board.length&&y>=0&&y<board[0].length&&!visited[x+1][y]&&board[x+1][y]==word.charAt(index)){
            visited[x+1][y]=true;
            if(dfs(board,visited,word,index+1,x+1,y)) return true;
            visited[x+1][y]=false;
        }
        if(x>=0&&x<board.length&&y-1>=0&&y-1<board[0].length&&!visited[x][y-1]&&board[x][y-1]==word.charAt(index)){
            visited[x][y-1]=true;
            if(dfs(board,visited,word,index+1,x,y-1)) return true;
            visited[x][y-1]=false;
        }
        if(x-1>=0&&x-1<board.length&&y>=0&&y<board[0].length&&!visited[x-1][y]&&board[x-1][y]==word.charAt(index)){
            visited[x-1][y]=true;
            if(dfs(board,visited,word,index+1,x-1,y)) return true;
            visited[x-1][y]=false;
        }
        return false;
    }
}