给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
单词必须按照字母顺序，通过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。
示例 1：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" 输出：true
示例 2：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" 输出：true
示例 3：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" 输出：false
提示：
m == board.lengthn = board[i].length1 <= m, n <= 61 <= word.length <= 15board和word仅由大小写英文字母组成
进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
回溯（深度优先搜索 DFS ）实现。
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
def dfs(i, j, cur):
if cur == len(word):
return True
if (
i < 0
or i >= m
or j < 0
or j >= n
or board[i][j] == '0'
or word[cur] != board[i][j]
):
return False
t = board[i][j]
board[i][j] = '0'
for a, b in [[0, 1], [0, -1], [-1, 0], [1, 0]]:
x, y = i + a, j + b
if dfs(x, y, cur + 1):
return True
board[i][j] = t
return False
m, n = len(board), len(board[0])
return any(dfs(i, j, 0) for i in range(m) for j in range(n))class Solution {
public boolean exist(char[][] board, String word) {
int m = board.length;
int n = board[0].length;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (dfs(i, j, 0, m, n, board, word)) {
return true;
}
}
}
return false;
}
private boolean dfs(int i, int j, int cur, int m, int n, char[][] board, String word) {
if (cur == word.length()) {
return true;
}
if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word.charAt(cur)) {
return false;
}
board[i][j] += 256;
int[] dirs = {-1, 0, 1, 0, -1};
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k];
int y = j + dirs[k + 1];
if (dfs(x, y, cur + 1, m, n, board, word)) {
return true;
}
}
board[i][j] -= 256;
return false;
}
}function exist(board: string[][], word: string): boolean {
let m = board.length,
n = board[0].length;
let visited = Array.from({ length: m }, v => new Array(n).fill(false));
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (dfs(board, word, i, j, 0, visited)) {
return true;
}
}
}
return false;
}
function dfs(
board: string[][],
word: string,
i: number,
j: number,
depth: number,
visited: boolean[][],
): boolean {
let m = board.length,
n = board[0].length;
if (i < 0 || i > m - 1 || j < 0 || j > n - 1 || visited[i][j]) {
return false;
}
if (board[i][j] != word.charAt(depth)) {
return false;
}
if (depth == word.length - 1) {
return true;
}
visited[i][j] = true;
++depth;
let res = false;
for (let [dx, dy] of [
[0, 1],
[0, -1],
[1, 0],
[-1, 0],
]) {
let x = i + dx,
y = j + dy;
res = res || dfs(board, word, x, y, depth, visited);
}
visited[i][j] = false;
return res;
}class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
int m = board.size(), n = board[0].size();
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
if (dfs(i, j, 0, m, n, board, word))
return true;
return false;
}
bool dfs(int i, int j, int cur, int m, int n, vector<vector<char>>& board, string& word) {
if (cur == word.size()) return true;
if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[cur]) return false;
char t = board[i][j];
board[i][j] = '0';
vector<int> dirs = {-1, 0, 1, 0, -1};
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (dfs(x, y, cur + 1, m, n, board, word)) return true;
}
board[i][j] = t;
return false;
}
};public class Solution {
public bool Exist(char[][] board, string word) {
var lenI = board.Length;
var lenJ = lenI == 0 ? 0 : board[0].Length;
var visited = new bool[lenI, lenJ];
for (var i = 0; i < lenI; ++i)
{
for (var j = 0; j < lenJ; ++j)
{
if (Search(board, visited, word, lenI, lenJ, i, j, 0))
{
return true;
}
}
}
return false;
}
private int[,] paths = new int[4,2] { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };
private bool Search(char[][] board, bool[,] visited, string word, int lenI, int lenJ, int i, int j, int p)
{
if (p == word.Length)
{
return true;
}
if (i < 0 || i >= lenI || j < 0 || j >= lenJ) return false;
if (visited[i, j] || word[p] != board[i][j]) return false;
visited[i, j] = true;
for (var k = 0; k < 4; ++k)
{
if (Search(board, visited, word, lenI, lenJ, i + paths[k, 0], j + paths[k, 1], p + 1)) return true;
}
visited[i, j] = false;
return false;
}
}func exist(board [][]byte, word string) bool {
m, n := len(board), len(board[0])
var dfs func(i, j, cur int) bool
dfs = func(i, j, cur int) bool {
if cur == len(word) {
return true
}
if i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[cur] {
return false
}
t := board[i][j]
board[i][j] = '0'
dirs := []int{-1, 0, 1, 0, -1}
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if dfs(x, y, cur+1) {
return true
}
}
board[i][j] = t
return false
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if dfs(i, j, 0) {
return true
}
}
}
return false
}impl Solution {
fn dfs(
i: usize,
j: usize,
c: usize,
word: &[u8],
board: &Vec<Vec<char>>,
vis: &mut Vec<Vec<bool>>,
) -> bool {
if board[i][j] as u8 != word[c] {
return false;
}
if c == word.len() - 1 {
return true;
}
vis[i][j] = true;
let dirs = [[-1, 0], [0, -1], [1, 0], [0, 1]];
for [x, y] in dirs.into_iter() {
let i = x + i as i32;
let j = y + j as i32;
if i < 0 || i == board.len() as i32 || j < 0 || j == board[0].len() as i32 {
continue;
}
let (i, j) = (i as usize, j as usize);
if !vis[i][j] && Self::dfs(i, j, c + 1, word, board, vis) {
return true;
}
}
vis[i][j] = false;
false
}
pub fn exist(board: Vec<Vec<char>>, word: String) -> bool {
let m = board.len();
let n = board[0].len();
let word = word.as_bytes();
let mut vis = vec![vec![false; n]; m];
for i in 0..m {
for j in 0..n {
if Self::dfs(i, j, 0, word, &board, &mut vis) {
return true;
}
}
}
false
}
}