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79. Word Search

Difficulty: Medium

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.

Example:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

Solution

Language: Java

class Solution {
    public boolean exist(char[][] board, String word) {
        if (board == null || board.length == 0) {
            return false;
        }
        int rows = board.length;
        int width = board[0].length;
        int[][] occupied = new int[rows][width];
        char[] chars = word.toCharArray();
        char firstChar = chars[0];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < width; j++) {
                if (board[i][j] == firstChar) {
                    occupied[i][j] = 1;
                    if (existSubStr(board, occupied, chars, 1, i, j)) {
                        return true;
                    } else {
                        occupied[i][j] = 0;
                    }
                }
            }
        }
        return false;
    }

    private boolean existSubStr(char[][] board, int[][] occupied, char[] chars, int startIndex, int i, int j) {
        if (startIndex >= chars.length) {
            return true;
        }
        char aChar = chars[startIndex];
        if (i + 1 < board.length && board[i + 1][j] == aChar && occupied[i + 1][j] == 0) {
            occupied[i + 1][j] = 1;
            if (existSubStr(board, occupied, chars, startIndex + 1, i + 1, j)) {
                return true;
            } else {
                occupied[i + 1][j] = 0;
            }
        }

        if (i - 1 >= 0 && board[i - 1][j] == aChar && occupied[i - 1][j] == 0) {
            occupied[i - 1][j] = 1;
            if (existSubStr(board, occupied, chars, startIndex + 1, i - 1, j)) {
                return true;
            } else {
                occupied[i - 1][j] = 0;
            }
        }


        if (j + 1 < board[0].length && board[i][j + 1] == aChar && occupied[i][j + 1] == 0) {
            occupied[i][j + 1] = 1;
            if (existSubStr(board, occupied, chars, startIndex + 1, i, j + 1)) {
                return true;
            } else {
                occupied[i][j + 1] = 0;
            }
        }

        if (j - 1 >= 0 && board[i][j - 1] == aChar && occupied[i][j - 1] == 0) {
            occupied[i][j - 1] = 1;
            if (existSubStr(board, occupied, chars, startIndex + 1, i, j - 1)) {
                return true;
            } else {
                occupied[i][j - 1] = 0;
            }
        }
        return false;
    }
}

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