Word Search

本文介绍了一个使用深度优先搜索(DFS)算法解决二维网格中查找指定单词的问题。通过递归方式检查相邻单元格来构建目标单词,并确保同一字母单元格不被重复使用。

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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.


Analysis: DFS. First, try to find the element that can be used as the start points. Then, for each such element, recursively search the elements around it (i.e., up, down, left, right). 

public class Solution {
    public boolean[][] flag = null;     // mark array, default initialized to be false
    
    public boolean exist(char[][] board, String word, int level, int lastRow, int lastCol) {
        if(level==word.length()) return true;
        
        // first invoked function, try to find the possible start points
        if(lastRow==-1 && lastCol==-1) {
            for(int i=0; i<board.length; i++) {
                for(int j=0; j<board[0].length; j++) {
                    if(board[i][j] == word.charAt(level)) {
                        flag[i][j] = true;
                        if(exist(board, word, level+1, i, j)) return true;
                        flag[i][j] = false;
                    }
                }
            }
        }
        // other invoked functions
        else {
            return checkAround(board, word, level+1, lastRow-1, lastCol) |      // up
                   checkAround(board, word, level+1, lastRow+1, lastCol) |      // down
                   checkAround(board, word, level+1, lastRow, lastCol-1) |      // left
                   checkAround(board, word, level+1, lastRow, lastCol+1);       // right
        }
        return false;
    }
    
    public boolean checkAround(char[][] board, String word, int level, int row, int col) {
        if(row>=0 && row<board.length && col>=0 && col<board[0].length && board[row][col]==word.charAt(level-1) && !flag[row][col]) {
            flag[row][col] = true;
            if(exist(board, word, level,row, col)) return true;
            flag[row][col] = false;
        }
        return false;
    }
    
    public boolean exist(char[][] board, String word) {
        if(board.length==0 || board[0].length==0 || word.length()==0) return false;
        flag = new boolean[board.length][board[0].length];
        return exist(board, word, 0, -1, -1);
    }
}

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