| comments | difficulty | edit_url |
|---|---|---|
true |
中等 |
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。
例如，在下面的 3×4 的矩阵中包含单词 "ABCCED"（单词中的字母已标出）。
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","d"]], word = "abcd" 输出：false
提示：
1 <= board.length <= 2001 <= board[i].length <= 200board和word仅由大小写英文字母组成
注意：本题与主站 79 题相同：https://leetcode.cn/problems/word-search/
我们可以枚举矩阵的每个位置 word 的路径。如果找到了一条路径，即可返回 true，否则在枚举完所有的位置后，返回 false。
那么问题的转换为如何采用深度优先搜索的方法寻找字符串 word 的路径。我们可以设计一个函数 word[k] 的情况下，是否能够找到字符串 word 的路径。如果能找到，返回 true，否则返回 false。
函数
- 如果当前字符
word[k]已经匹配到字符串word的末尾，说明已经找到了字符串word的路径，返回true。 - 如果当前位置
$(i, j)$ 超出矩阵边界，或者当前位置的字符与word[k]不同，说明当前位置不在字符串word的路径上，返回false。 - 否则，我们将当前位置的字符标记为已访问（防止重复搜索），然后分别向当前位置的上、下、左、右四个方向继续匹配字符
word[k + 1]，只要有一条路径能够匹配到字符串word的路径，就说明能够找到字符串word的路径，返回true。在回溯时，我们要将当前位置的字符还原为未访问过的状态。
时间复杂度 word 的长度。我们需要枚举矩阵中的每个位置，然后对于每个位置，我们最多需要搜索三个方向。
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
def dfs(i, j, k):
if k == len(word):
return True
if i < 0 or i >= m or j < 0 or j >= n or board[i][j] != word[k]:
return False
board[i][j] = ""
dirs = (-1, 0, 1, 0, -1)
ans = any(dfs(i + a, j + b, k + 1) for a, b in pairwise(dirs))
board[i][j] = word[k]
return ans
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 {
private char[][] board;
private String word;
private int m;
private int n;
public boolean exist(char[][] board, String word) {
m = board.length;
n = board[0].length;
this.board = board;
this.word = word;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (dfs(i, j, 0)) {
return true;
}
}
}
return false;
}
private boolean dfs(int i, int j, int k) {
if (k == word.length()) {
return true;
}
if (i < 0 || i >= m || j < 0 || j >= n || word.charAt(k) != board[i][j]) {
return false;
}
board[i][j] = ' ';
int[] dirs = {-1, 0, 1, 0, -1};
boolean ans = false;
for (int l = 0; l < 4; ++l) {
ans = ans || dfs(i + dirs[l], j + dirs[l + 1], k + 1);
}
board[i][j] = word.charAt(k);
return ans;
}
}class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
int m = board.size(), n = board[0].size();
int dirs[5] = {-1, 0, 1, 0, -1};
function<bool(int, int, int)> dfs = [&](int i, int j, int k) -> bool {
if (k == word.size()) {
return true;
}
if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[k]) {
return false;
}
board[i][j] = '.';
bool ans = 0;
for (int l = 0; l < 4; ++l) {
ans |= dfs(i + dirs[l], j + dirs[l + 1], k + 1);
}
board[i][j] = word[k];
return ans;
};
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (dfs(i, j, 0)) {
return true;
}
}
}
return false;
}
};func exist(board [][]byte, word string) bool {
m, n := len(board), len(board[0])
var dfs func(i, j, k int) bool
dfs = func(i, j, k int) bool {
if k == len(word) {
return true
}
if i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[k] {
return false
}
board[i][j] = ' '
dirs := []int{-1, 0, 1, 0, -1}
ans := false
for l := 0; l < 4; l++ {
ans = ans || dfs(i+dirs[l], j+dirs[l+1], k+1)
}
board[i][j] = word[k]
return ans
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if dfs(i, j, 0) {
return true
}
}
}
return false
}function exist(board: string[][], word: string): boolean {
const m = board.length;
const n = board[0].length;
const dfs = (i: number, j: number, k: number) => {
if ((board[i] ?? [])[j] !== word[k]) {
return false;
}
if (++k === word.length) {
return true;
}
const temp = board[i][j];
board[i][j] = ' ';
if (dfs(i + 1, j, k) || dfs(i, j + 1, k) || dfs(i - 1, j, k) || dfs(i, j - 1, k)) {
return true;
}
board[i][j] = temp;
return false;
};
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
if (dfs(i, j, 0)) {
return true;
}
}
}
return false;
}impl Solution {
fn dfs(
board: &mut Vec<Vec<char>>,
chars: &Vec<char>,
i: usize,
j: usize,
mut k: usize,
) -> bool {
if board[i][j] != chars[k] {
return false;
}
k += 1;
if k == chars.len() {
return true;
}
let temp = board[i][j];
board[i][j] = ' ';
if (i != 0 && Self::dfs(board, chars, i - 1, j, k))
|| (j != 0 && Self::dfs(board, chars, i, j - 1, k))
|| (i != board.len() - 1 && Self::dfs(board, chars, i + 1, j, k))
|| (j != board[0].len() - 1 && Self::dfs(board, chars, i, j + 1, k))
{
return true;
}
board[i][j] = temp;
false
}
pub fn exist(mut board: Vec<Vec<char>>, word: String) -> bool {
let m = board.len();
let n = board[0].len();
let chars = word.chars().collect::<Vec<char>>();
for i in 0..m {
for j in 0..n {
if Self::dfs(&mut board, &chars, i, j, 0) {
return true;
}
}
}
false
}
}/**
* @param {character[][]} board
* @param {string} word
* @return {boolean}
*/
var exist = function (board, word) {
const m = board.length;
const n = board[0].length;
const dirs = [-1, 0, 1, 0, -1];
const dfs = (i, j, k) => {
if (k == word.length) {
return true;
}
if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[k]) {
return false;
}
board[i][j] = ' ';
let ans = false;
for (let l = 0; l < 4; ++l) {
ans = ans || dfs(i + dirs[l], j + dirs[l + 1], k + 1);
}
board[i][j] = word[k];
return ans;
};
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (dfs(i, j, 0)) {
return true;
}
}
}
return false;
};public class Solution {
private char[][] board;
private string word;
private int m;
private int n;
public bool Exist(char[][] board, string word) {
m = board.Length;
n = board[0].Length;
this.board = board;
this.word = word;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (dfs(i, j, 0)) {
return true;
}
}
}
return false;
}
private bool dfs(int i, int j, int k) {
if (k == word.Length) {
return true;
}
if (i < 0 || i >= m || j < 0 || j >= n || word[k] != board[i][j]) {
return false;
}
board[i][j] = ' ';
int[] dirs = {-1, 0, 1, 0, -1};
bool ans = false;
for (int l = 0; l < 4; ++l) {
ans = ans || dfs(i + dirs[l], j + dirs[l + 1], k + 1);
}
board[i][j] = word[k];
return ans;
}
}class Solution {
private var board: [[Character]]
private var word: String
private var m: Int
private var n: Int
init() {
self.board = []
self.word = ""
self.m = 0
self.n = 0
}
func exist(_ board: [[Character]], _ word: String) -> Bool {
self.board = board
self.word = word
m = board.count
n = board[0].count
for i in 0..<m {
for j in 0..<n {
if dfs(i, j, 0) {
return true
}
}
}
return false
}
private func dfs(_ i: Int, _ j: Int, _ k: Int) -> Bool {
if k == word.count {
return true
}
if i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[word.index(word.startIndex, offsetBy: k)] {
return false
}
let temp = board[i][j]
board[i][j] = " "
let dirs = [-1, 0, 1, 0, -1]
var ans = false
for l in 0..<4 {
let ni = i + dirs[l]
let nj = j + dirs[l + 1]
if dfs(ni, nj, k + 1) {
ans = true
break
}
}
board[i][j] = temp
return ans
}
}