-
Notifications
You must be signed in to change notification settings - Fork 2
/
剑指Offer12.矩阵中的路径.html
89 lines (76 loc) · 2.79 KB
/
剑指Offer12.矩阵中的路径.html
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<meta http-equiv="X-UA-Compatible" content="IE=edge" />
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<title>剑指 Offer 12. 矩阵中的路径</title>
</head>
<body>
<script>
// https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof/
// 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。
// 提示:
// 1 <= board.length <= 200
// 1 <= board[i].length <= 200
// board 和 word 仅由大小写英文字母组成
/**
* @param {character[][]} board
* @param {string} word
* @return {boolean}
*/
function find(board, i, j, word, idx) {
// 越界或字符不对
if (i < 0 || i > board.length - 1 || j < 0 || j > board[0].length - 1 || board[i][j] !== word[idx])
return false;
// 查找结束
if (idx === word.length - 1) return true;
// 已访问过
board[i][j] = 'vis';
// 从四个方向进行遍历
let res =
find(board, i + 1, j, word, idx + 1) ||
find(board, i - 1, j, word, idx + 1) ||
find(board, i, j + 1, word, idx + 1) ||
find(board, i, j - 1, word, idx + 1);
// 恢复或者也可以说回溯, 自身做visited数组,降低空间复杂度和逻辑复杂度
board[i][j] = word[idx];
return res;
}
// M,N 分别为矩阵行列大小, KK 为字符串 word 长度。
// 时间复杂度 O(3^KMN) 空间复杂度 O(K) 搜索过程中的递归深度不超过 K
var exist = function (board, word) {
for (let i = 0; i < board.length; i++) {
for (let j = 0; j < board[0].length; j++) {
if (find(board, i, j, word, 0)) {
return true;
}
}
}
return false;
};
// --- answer-1 ---
// --- answer-1 ---
// --- answer-2 ---
// --- answer-2 ---
var board = [
['A', 'B', 'C', 'E'],
['S', 'F', 'C', 'S'],
['A', 'D', 'E', 'E']
],
word = 'ABCCED';
var result = true;
var board = [
['a', 'b'],
['c', 'd']
],
word = 'abcd';
var result = false;
console.log('board = ', board);
console.log('word = ', word);
console.log('result = ', result);
console.log('exist = ', exist(board, word));
</script>
</body>
</html>