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.
DFS + Backtracking. Replace the cell with NaN before
proceeding to the next level and restore when backtracking.
/**
* @param {character[][]} board
* @param {string} word
* @return {boolean}
*/
let exist = function (board, word) {
const height = board.length;
if (height <= 0) {
return false;
}
const width = board[0].length;
if (width <= 0) {
return false;
}
for (let row = 0; row < height; row++) {
for (let col = 0; col < width; col++) {
if (
board[row][col] === word[0] &&
_exist(
board,
word,
0,
[
[-1, 0],
[1, 0],
[0, -1],
[0, 1],
],
row,
col
)
) {
return true;
}
}
}
return false;
};
function _exist(board, word, iWord, directions, row, col) {
if (iWord === word.length) {
return true;
}
if (!board[row] || word[iWord] !== board[row][col]) {
return false;
}
const cell = board[row][col];
board[row][col] = NaN;
for (let i = directions.length - 1; i >= 0; i--) {
if (
_exist(
board,
word,
iWord + 1,
directions,
row + directions[i][0],
col + directions[i][1]
)
) {
return true;
}
}
board[row][col] = cell;
return false;
}☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆
☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆