本问题对应的 leetcode 原文链接：剑指 Offer 12. 矩阵中的路径
问题描述
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
限制：
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
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
int n;
int m;
int len;
boolean[][] visited;
public boolean exist(char[][] board, String word) {
this.n = board.length;
this.m = board[0].length;
this.len = word.length();
visited = new boolean[n][m];
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(dsf(board, i, j, word, 0)){
return true;
}
}
}
return false;
}
boolean dsf(char[][] board, int i, int j, String word, int k){
// 判断是否越界，已经访问过或者不匹配
if(i < 0 || i >= n || j < 0 || j >= m || board[i][j] != word.charAt(k)){
return false;
}
if(k == len - 1){
return true;
}
//visited[i][j] = true;
board[i][j] = '\n';
// 四个方向都搜索一下
boolean res = dsf(board, i, j + 1, word, k + 1) ||
dsf(board, i + 1, j, word, k + 1) ||
dsf(board, i, j - 1, word, k + 1) ||
dsf(board, i - 1, j, word, k + 1);
board[i][j] = word.charAt(k);
return res;
// 时间复杂度：mn * 3^len
}
}
时间复杂度：O(3^K * MN) ： 最差情况下，需要遍历矩阵中长度为 KK 字符串的所有方案，时间复杂度为 O(3^K)；矩阵中共有 MN 个起点，时间复杂度为 O(MN)
空间复杂度：O(k)
Python
class Solution(object):
def exist(self, board, word):
"""
:type board: List[List[str]]
:type word: str
:rtype: bool
"""
n = len(board)
m = len(board[0])
l = len(word)
visited = [[False] * m for _ in range(n)]
def dfs(i, j, k):
if i < 0 or i >= n or j < 0 or j >= m or visited[i][j] or board[i][j] != word[k]:
return False
if k == l - 1:
return True
visited[i][j] = True
res = dfs(i, j + 1, k + 1) or dfs(i + 1, j, k + 1) or dfs(i, j - 1, k + 1) or dfs(i - 1, j, k + 1)
visited[i][j] = False
return res
for i in range(n):
for j in range(m):
if dfs(i, j, 0):
return True
return False
C++
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
int n = board.size();
int m = board[0].size();
int l = word.size();
vector<vector<bool>> visited(n, vector<bool>(m, false));
function<bool(int, int, int)> dfs = [&](int i, int j, int k) -> bool {
if (i < 0 || i >= n || j < 0 || j >= m || visited[i][j] || board[i][j] != word[k]) {
return false;
}
if (k == l - 1) {
return true;
}
visited[i][j] = true;
bool res = dfs(i, j + 1, k + 1) || dfs(i + 1, j, k + 1) || dfs(i, j - 1, k + 1) || dfs(i - 1, j, k + 1);
visited[i][j] = false;
return res;
};
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (dfs(i, j, 0)) {
return true;
}
}
}
return false;
}
};
Go
func exist(board [][]byte, word string) bool {
n := len(board)
m := len(board[0])
l := len(word)
visited := make([][]bool, n)
for i := range visited {
visited[i] = make([]bool, m)
}
var dfs func(int, int, int) bool
dfs = func(i, j, k int) bool {
if i < 0 || i >= n || j < 0 || j >= m || visited[i][j] || board[i][j] != word[k] {
return false
}
if k == l-1 {
return true
}
visited[i][j] = true
res := dfs(i, j+1, k+1) || dfs(i+1, j, k+1) || dfs(i, j-1, k+1) || dfs(i-1, j, k+1)
visited[i][j] = false
return res
}
for i := 0; i < n; i++ {
for j := 0; j < m; j++ {
if dfs(i, j, 0) {
return true
}
}
}
return false
}
JS
/**
* @param {character[][]} board
* @param {string} word
* @return {boolean}
*/
var exist = function(board, word) {
// 初始化变量
let n = board.length;
let m = board[0].length;
let len = word.length;
let visited = new Array(n).fill().map(() => new Array(m).fill(false));
// 开始搜索
for(let i = 0; i < n; i++){
for(let j = 0; j < m; j++){
if(dsf(board, i, j, word, 0)){
return true;
}
}
}
return false;
};
function dsf(board, i, j, word, k) {
// 判断是否越界，已经访问过或者不匹配
if(i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] !== word.charAt(k)){
return false;
}
if(k === word.length - 1){
return true;
}
board[i][j] = '\n';
// 四个方向都搜索一下
let res = dsf(board, i, j + 1, word, k + 1) ||
dsf(board, i + 1, j, word, k + 1) ||
dsf(board, i, j - 1, word, k + 1) ||
dsf(board, i - 1, j, word, k + 1);
board[i][j] = word.charAt(k);
return res;
// 时间复杂度：mn * 3^len
}
shuaidi