79. Word Search

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 =

[
[‘A’,’B’,’C’,’E’],
[‘S’,’F’,’C’,’S’],
[‘A’,’D’,’E’,’E’]
]
word = “ABCCED”, -> returns true,
word = “SEE”, -> returns true,
word = “ABCB”, -> returns false.

典型的DFS问题

      public static boolean exist(char[][] board, String word) {  
        // 对每一个节点进行深搜  
        for(int i=0; i<board.length; i++){  
            for(int j=0; j<board[0].length; j++){  
                if(dfs(board, word, 0, i, j)){  
                    return true;  
                }  
            }  
        }  
        return false;  
    }  

    // dfs搜索  
    public static boolean dfs(char[][] board, String word, int index, int x, int y){  
        if(index == word.length()-1 && word.charAt(index)==board[x][y]){  
            return true;  
        }  

        if(word.charAt(index) != board[x][y]){  
            return false;  
        }  

        char tmp = board[x][y];     // 保存原始值  
        board[x][y] = '.';  
        boolean b1 = false, b2 = false, b3 = false, b4 = false;  
        if(x-1 >= 0 && board[x-1][y] != '.'){  
            b1 = dfs(board, word, index+1, x-1, y);  
        }  
        if(!b1 && y-1>=0 && board[x][y-1] != '.'){  
            b2 = dfs(board, word, index+1, x, y-1);  
        }  
        if(!b1 && !b2 && x+1<board.length && board[x+1][y] != '.'){  
            b3 = dfs(board, word, index+1, x+1, y);  
        }  
        if(!b1 && !b2 && !b3 && y+1<board[0].length && board[x][y+1] != '.'){  
            b4 = dfs(board, word, index+1, x, y+1);  
        }  

        board[x][y] = tmp;          // 还原原始值  
        return b1 || b2 || b3 || b4;  
    }