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Commit 8a2235a

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feat: update solutions to lc problem: No.0688 (doocs#3845)
1 parent 3057d04 commit 8a2235a

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‎solution/0600-0699/0688.Knight Probability in Chessboard/README.md‎

Lines changed: 29 additions & 45 deletions
Original file line numberDiff line numberDiff line change
@@ -65,19 +65,19 @@ tags:
6565

6666
### 方法一:动态规划
6767

68-
我们定义 $f[h][i][j]$ 表示骑士从 $(i, j)$ 位置出发,走了 $h$ 步以后还留在棋盘上的概率。那么最终答案就是 $f[k][row][column]$。
68+
我们定义 $f[h][i][j]$ 表示骑士从 $(i, j)$ 位置出发,走了 $h$ 步以后还留在棋盘上的概率。那么最终答案就是 $f[k][\textit{row}][\textit{column}]$。
6969

7070
当 $h=0$ 时,骑士一定在棋盘上,概率为 1ドル,ドル即 $f[0][i][j]=1$。
7171

7272
当 $h \gt 0$ 时,骑士在 $(i, j)$ 位置上的概率可以由其上一步的 8ドル$ 个位置上的概率转移得到,即:
7373

7474
$$
75-
f[h][i][j] = \sum_{a, b} f[h - 1][a][b] \times \frac{1}{8}
75+
f[h][i][j] = \sum_{x, y} f[h - 1][x][y] \times \frac{1}{8}
7676
$$
7777

78-
其中 $(a, b)$ 是从 $(i, j)$ 位置可以走到的 8ドル$ 个位置中的一个。
78+
其中 $(x, y)$ 是从 $(i, j)$ 位置可以走到的 8ドル$ 个位置中的一个。
7979

80-
最终答案即为 $f[k][row][column]$。
80+
最终答案即为 $f[k][\textit{row}][\textit{column}]$。
8181

8282
时间复杂度 $O(k \times n^2),ドル空间复杂度 $O(k \times n^2)$。其中 $k$ 和 $n$ 分别是给定的步数和棋盘大小。
8383

@@ -197,9 +197,9 @@ func knightProbability(n int, k int, row int, column int) float64 {
197197

198198
```ts
199199
function knightProbability(n: number, k: number, row: number, column: number): number {
200-
const f = newArray(k + 1)
201-
.fill(0)
202-
.map(() =>newArray(n).fill(0).map(() =>newArray(n).fill(0)));
200+
const f = Array.from({ length: k + 1 }, () =>
201+
Array.from({ length: n }, () =>Array(n).fill(0)),
202+
);
203203
for (let i = 0; i < n; ++i) {
204204
for (let j = 0; j < n; ++j) {
205205
f[0][i][j] = 1;
@@ -226,55 +226,39 @@ function knightProbability(n: number, k: number, row: number, column: number): n
226226
#### Rust
227227

228228
```rust
229-
const DIR: [(i32, i32); 8] = [
230-
(-2, -1),
231-
(2, -1),
232-
(-1, -2),
233-
(1, -2),
234-
(2, 1),
235-
(-2, 1),
236-
(1, 2),
237-
(-1, 2),
238-
];
239-
const P: f64 = 1.0 / 8.0;
240-
241229
impl Solution {
242-
#[allow(dead_code)]
243230
pub fn knight_probability(n: i32, k: i32, row: i32, column: i32) -> f64 {
244-
// Here dp[i][j][k] represents through `i` steps, the probability that the knight stays on the board
245-
// Starts from row: `j`, column: `k`
246-
let mut dp: Vec<Vec<Vec<f64>>> =
247-
vec![vec![vec![0 as f64; n as usize]; n as usize]; k as usize + 1];
248-
249-
// Initialize the dp vector, since dp[0][j][k] should be 1
250-
for j in 0..n as usize {
251-
for k in 0..n as usize {
252-
dp[0][j][k] = 1.0;
231+
let n = n as usize;
232+
let k = k as usize;
233+
234+
let mut f = vec![vec![vec![0.0; n]; n]; k + 1];
235+
236+
for i in 0..n {
237+
for j in 0..n {
238+
f[0][i][j] = 1.0;
253239
}
254240
}
255241

256-
// Begin the actual dp process
257-
for i in 1..=k {
258-
for j in 0..n {
259-
for k in 0..n {
260-
for (dx, dy) in DIR {
261-
let x = j + dx;
262-
let y = k + dy;
263-
if Self::check_bounds(x, y, n, n) {
264-
dp[i as usize][j as usize][k as usize] +=
265-
P * dp[(i as usize) - 1][x as usize][y as usize];
242+
let dirs = [-2, -1, 2, 1, -2, 1, 2, -1, -2];
243+
244+
for h in 1..=k {
245+
for i in 0..n {
246+
for j in 0..n {
247+
for p in 0..8 {
248+
let x = i as isize + dirs[p];
249+
let y = j as isize + dirs[p + 1];
250+
251+
if x >= 0 && x < n as isize && y >= 0 && y < n as isize {
252+
let x = x as usize;
253+
let y = y as usize;
254+
f[h][i][j] += f[h - 1][x][y] / 8.0;
266255
}
267256
}
268257
}
269258
}
270259
}
271260

272-
dp[k as usize][row as usize][column as usize]
273-
}
274-
275-
#[allow(dead_code)]
276-
fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
277-
i >= 0 && i < n && j >= 0 && j < m
261+
f[k][row as usize][column as usize]
278262
}
279263
}
280264
```

‎solution/0600-0699/0688.Knight Probability in Chessboard/README_EN.md‎

Lines changed: 32 additions & 48 deletions
Original file line numberDiff line numberDiff line change
@@ -61,21 +61,21 @@ The total probability the knight stays on the board is 0.0625.
6161

6262
### Solution 1: Dynamic Programming
6363

64-
Let $f[h][i][j]$ denotes the probability that the knight is still on the chessboard after $h$ steps starting from the position $(i, j)$. Then the final answer is $f[k][row][column]$.
64+
We define $f[h][i][j]$ to represent the probability that the knight remains on the board after taking $h$ steps starting from position $(i, j)$. The final answer is $f[k][\textit{row}][\textit{column}]$.
6565

66-
When $h = 0,ドル the knight is always on the chessboard, so $f[0][i][j] = 1$.
66+
When $h=0,ドル the knight is definitely on the board, so the probability is 1ドル,ドル i.e., $f[0][i][j]=1$.
6767

68-
When $h \gt 0,ドル the probability that the knight is on the position $(i, j)$ can be transferred from the probability on its 8ドル$ adjacent positions, which are:
68+
When $h \gt 0,ドル the probability that the knight is at position $(i, j)$ can be derived from the probabilities of the 8ドル$ possible positions it could have come from in the previous step, i.e.,
6969

7070
$$
71-
f[h][i][j] = \sum_{a, b} f[h - 1][a][b] \times \frac{1}{8}
71+
f[h][i][j] = \sum_{x, y} f[h - 1][x][y] \times \frac{1}{8}
7272
$$
7373

74-
where $(a, b)$ is one of the 8ドル$ adjacent positions.
74+
where $(x, y)$ is one of the 8ドル$ positions the knight can move to from $(i, j)$.
7575

76-
The final answer is $f[k][row][column]$.
76+
The final answer is $f[k][\textit{row}][\textit{column}]$.
7777

78-
The time complexity is $O(k \times n^2),ドル and the space complexity is $O(k \times n^2)$. Here $k$ and $n$ are the given steps and the chessboard size, respectively.
78+
The time complexity is $O(k \times n^2),ドル and the space complexity is $O(k \times n^2)$. Here, $k$ and $n$ are the given number of steps and the size of the board, respectively.
7979

8080
<!-- tabs:start -->
8181

@@ -193,9 +193,9 @@ func knightProbability(n int, k int, row int, column int) float64 {
193193

194194
```ts
195195
function knightProbability(n: number, k: number, row: number, column: number): number {
196-
const f = newArray(k + 1)
197-
.fill(0)
198-
.map(() =>newArray(n).fill(0).map(() =>newArray(n).fill(0)));
196+
const f = Array.from({ length: k + 1 }, () =>
197+
Array.from({ length: n }, () =>Array(n).fill(0)),
198+
);
199199
for (let i = 0; i < n; ++i) {
200200
for (let j = 0; j < n; ++j) {
201201
f[0][i][j] = 1;
@@ -222,55 +222,39 @@ function knightProbability(n: number, k: number, row: number, column: number): n
222222
#### Rust
223223

224224
```rust
225-
const DIR: [(i32, i32); 8] = [
226-
(-2, -1),
227-
(2, -1),
228-
(-1, -2),
229-
(1, -2),
230-
(2, 1),
231-
(-2, 1),
232-
(1, 2),
233-
(-1, 2),
234-
];
235-
const P: f64 = 1.0 / 8.0;
236-
237225
impl Solution {
238-
#[allow(dead_code)]
239226
pub fn knight_probability(n: i32, k: i32, row: i32, column: i32) -> f64 {
240-
// Here dp[i][j][k] represents through `i` steps, the probability that the knight stays on the board
241-
// Starts from row: `j`, column: `k`
242-
let mut dp: Vec<Vec<Vec<f64>>> =
243-
vec![vec![vec![0 as f64; n as usize]; n as usize]; k as usize + 1];
244-
245-
// Initialize the dp vector, since dp[0][j][k] should be 1
246-
for j in 0..n as usize {
247-
for k in 0..n as usize {
248-
dp[0][j][k] = 1.0;
227+
let n = n as usize;
228+
let k = k as usize;
229+
230+
let mut f = vec![vec![vec![0.0; n]; n]; k + 1];
231+
232+
for i in 0..n {
233+
for j in 0..n {
234+
f[0][i][j] = 1.0;
249235
}
250236
}
251237

252-
// Begin the actual dp process
253-
for i in 1..=k {
254-
for j in 0..n {
255-
for k in 0..n {
256-
for (dx, dy) in DIR {
257-
let x = j + dx;
258-
let y = k + dy;
259-
if Self::check_bounds(x, y, n, n) {
260-
dp[i as usize][j as usize][k as usize] +=
261-
P * dp[(i as usize) - 1][x as usize][y as usize];
238+
let dirs = [-2, -1, 2, 1, -2, 1, 2, -1, -2];
239+
240+
for h in 1..=k {
241+
for i in 0..n {
242+
for j in 0..n {
243+
for p in 0..8 {
244+
let x = i as isize + dirs[p];
245+
let y = j as isize + dirs[p + 1];
246+
247+
if x >= 0 && x < n as isize && y >= 0 && y < n as isize {
248+
let x = x as usize;
249+
let y = y as usize;
250+
f[h][i][j] += f[h - 1][x][y] / 8.0;
262251
}
263252
}
264253
}
265254
}
266255
}
267256

268-
dp[k as usize][row as usize][column as usize]
269-
}
270-
271-
#[allow(dead_code)]
272-
fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
273-
i >= 0 && i < n && j >= 0 && j < m
257+
f[k][row as usize][column as usize]
274258
}
275259
}
276260
```
Lines changed: 21 additions & 37 deletions
Original file line numberDiff line numberDiff line change
@@ -1,51 +1,35 @@
1-
const DIR: [(i32, i32); 8] = [
2-
(-2, -1),
3-
(2, -1),
4-
(-1, -2),
5-
(1, -2),
6-
(2, 1),
7-
(-2, 1),
8-
(1, 2),
9-
(-1, 2),
10-
];
11-
const P: f64 = 1.0 / 8.0;
12-
131
impl Solution {
14-
#[allow(dead_code)]
152
pub fn knight_probability(n: i32, k: i32, row: i32, column: i32) -> f64 {
16-
// Here dp[i][j][k] represents through `i` steps, the probability that the knight stays on the board
17-
// Starts from row: `j`, column: `k`
18-
letmut dp:Vec<Vec<Vec<f64>>> =
19-
vec![vec![vec![0asf64; nasusize]; nasusize]; kasusize + 1];
3+
let n = n asusize;
4+
let k = k asusize;
5+
6+
letmut f = vec![vec![vec![0.0; n]; n]; k + 1];
207

21-
// Initialize the dp vector, since dp[0][j][k] should be 1
22-
for j in 0..n as usize {
23-
for k in 0..n as usize {
24-
dp[0][j][k] = 1.0;
8+
for i in 0..n {
9+
for j in 0..n {
10+
f[0][i][j] = 1.0;
2511
}
2612
}
2713

28-
// Begin the actual dp process
29-
for i in 1..=k {
30-
for j in 0..n {
31-
for k in 0..n {
32-
for (dx, dy) in DIR {
33-
let x = j + dx;
34-
let y = k + dy;
35-
if Self::check_bounds(x, y, n, n) {
36-
dp[i as usize][j as usize][k as usize] +=
37-
P * dp[(i as usize) - 1][x as usize][y as usize];
14+
let dirs = [-2, -1, 2, 1, -2, 1, 2, -1, -2];
15+
16+
for h in 1..=k {
17+
for i in 0..n {
18+
for j in 0..n {
19+
for p in 0..8 {
20+
let x = i as isize + dirs[p];
21+
let y = j as isize + dirs[p + 1];
22+
23+
if x >= 0 && x < n as isize && y >= 0 && y < n as isize {
24+
let x = x as usize;
25+
let y = y as usize;
26+
f[h][i][j] += f[h - 1][x][y] / 8.0;
3827
}
3928
}
4029
}
4130
}
4231
}
4332

44-
dp[k as usize][row as usize][column as usize]
45-
}
46-
47-
#[allow(dead_code)]
48-
fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
49-
i >= 0 && i < n && j >= 0 && j < m
33+
f[k][row as usize][column as usize]
5034
}
5135
}

‎solution/0600-0699/0688.Knight Probability in Chessboard/Solution.ts‎

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -1,7 +1,7 @@
11
function knightProbability(n: number, k: number, row: number, column: number): number {
2-
const f = newArray(k + 1)
3-
.fill(0)
4-
.map(()=>newArray(n).fill(0).map(()=>newArray(n).fill(0)));
2+
const f = Array.from({length: k + 1},()=>
3+
Array.from({length: n},()=>Array(n).fill(0)),
4+
);
55
for (let i = 0; i < n; ++i) {
66
for (let j = 0; j < n; ++j) {
77
f[0][i][j] = 1;

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