079 Word Search 单词搜索

给定一个二维面板和一个单词,找出该单词是否存在于网格中。
这个词可由顺序相邻的单元格内的字母构成,其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
例如,
给定 二维面板 =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]
单词= "ABCCED",-> 返回 true,
单词 = "SEE", -> 返回 true,
单词 = "ABCB", -> 返回 false。
详见:https://leetcode.com/problems/word-search/description/

Java实现:

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

参考:https://www.cnblogs.com/springfor/p/3883942.html

posted on 2018-04-03 22:44  lina2014  阅读(188)  评论(0编辑  收藏  举报

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