-
Notifications
You must be signed in to change notification settings - Fork 157
Expand file tree
/
Copy path0529_minesweeper_(medium).js
More file actions
132 lines (122 loc) · 3.74 KB
/
Copy path0529_minesweeper_(medium).js
File metadata and controls
132 lines (122 loc) · 3.74 KB
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
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
/**
* 529. Minesweeper
*
* https://leetcode.com/problems/minesweeper/
*
* You are given a 2D char matrix representing the game board.
* 'M' represents an unrevealed mine,
* 'E' represents an unrevealed empty square,
* 'B' represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines,
* digit ('1' to '8') represents how many mines are adjacent to this revealed square,
* 'X' represents a revealed mine.
*
* Now given the next click position (row and column indices) among all the unrevealed squares ('M' or 'E'),
* return the board after revealing this position according to the following rules:
*
* - If a mine ('M') is revealed, then the game is over - change it to 'X'.
* - If an empty square ('E') with no adjacent mines is revealed,
* then change it to revealed blank ('B') and all of its adjacent unrevealed squares should be revealed recursively.
* - If an empty square ('E') with at least one adjacent mine is revealed, then change it to a digit ('1' to '8')
* representing the number of adjacent mines.
*
* Return the board when no more squares will be revealed.
*
* Example 1:
* Input:
*
* [['E', 'E', 'E', 'E', 'E'],
* ['E', 'E', 'M', 'E', 'E'],
* ['E', 'E', 'E', 'E', 'E'],
* ['E', 'E', 'E', 'E', 'E']]
*
* Click : [3,0]
*
* Output:
*
* [['B', '1', 'E', '1', 'B'],
* ['B', '1', 'M', '1', 'B'],
* ['B', '1', '1', '1', 'B'],
* ['B', 'B', 'B', 'B', 'B']]
*
* Example 2:
* Input:
*
* [['B', '1', 'E', '1', 'B'],
* ['B', '1', 'M', '1', 'B'],
* ['B', '1', '1', '1', 'B'],
* ['B', 'B', 'B', 'B', 'B']]
*
* Click : [1,2]
*
* Output:
*
* [['B', '1', 'E', '1', 'B'],
* ['B', '1', 'X', '1', 'B'],
* ['B', '1', '1', '1', 'B'],
* ['B', 'B', 'B', 'B', 'B']]
*
* Note:
* - The range of the input matrix's height and width is [1,50].
* - The click position will only be an unrevealed square ('M' or 'E'),
* which also means the input board contains at least one clickable square.
* - The input board won't be a stage when game is over (some mines have been revealed).
* - For simplicity, not mentioned rules should be ignored in this problem. For example,
* you don't need to reveal all the unrevealed mines when the game is over,
* consider any cases that you will win the game or flag any squares.
*/
/**
* @param {character[][]} board
* @param {number[]} click
* @return {character[][]}
*/
const updateBoard = (board, click) => {
const [x, y] = click;
if (board[x][y] === 'M') {
// If a mine ('M') is revealed, then the game is over - change it to 'X'
board[x][y] = 'X';
} else {
dfs(board, x, y);
}
return board;
};
const dfs = (board, x, y) => {
const dirs = [
[-1, -1],
[-1, 0],
[-1, 1],
[0, 1],
[1, 1],
[1, 0],
[1, -1],
[0, -1],
];
const m = board.length;
const n = board[0].length;
// Count the adjacent mines
let minesCount = 0;
for (let [dx, dy] of dirs) {
const i = x + dx;
const j = y + dy;
if (i >= 0 && i < m && j >= 0 && j < n && board[i][j] === 'M') {
minesCount++;
}
}
if (minesCount > 0) {
// If an empty square ('E') with at least one adjacent mine is revealed,
// then change it to a digit ('1' to '8') representing the number of adjacent mines.
board[x][y] = '' + minesCount;
} else {
// If an empty square ('E') with no adjacent mines is revealed, then change it to revealed blank ('B')
// and all of its adjacent unrevealed squares should be revealed recursively.
board[x][y] = 'B';
for (let [dx, dy] of dirs) {
const i = x + dx;
const j = y + dy;
if (i >= 0 && i < m && j >= 0 && j < n && board[i][j] === 'E') {
board[i][j] = 'B';
dfs(board, i, j);
}
}
}
};
export { updateBoard };