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.

Analysis

link: https://leetcode.com/problems/word-search/discuss/27811/My-Java-solution

Solution from Forum

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[i].length; j++){
            if((word.charAt(0) == board[i][j]) && search(board, word, i, j, 0, visited)){
                return true;
            }
        }
    }

    return false;
}

private boolean search(char[][]board, String word, int i, int j, int index, boolean[][] visited){
    if(index == word.length()){
        return true;
    }

    if(i >= board.length || i < 0 || j >= board[i].length || j < 0 || board[i][j] != word.charAt(index) || visited[i][j]){
        return false;
    }

    visited[i][j] = true;
    if(search(board, word, i-1, j, index+1, visited) || 
       search(board, word, i+1, j, index+1, visited) ||
       search(board, word, i, j-1, index+1, visited) || 
       search(board, word, i, j+1, index+1, visited)){
        return true;
    }

    visited[i][j] = false;
    return false;
}

Time complexity

m*n*4^k where k is the length of the string