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Commit 13bd762

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feat: add solutions to lc problems: No.1269,1275
* No.1269.Number of Ways to Stay in the Same Place After Some Steps * No.1275.Find Winner on a Tic Tac Toe Game
1 parent 46c8b91 commit 13bd762

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10 files changed

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-107
lines changed

10 files changed

+345
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‎solution/1200-1299/1269.Number of Ways to Stay in the Same Place After Some Steps/README.md‎

Lines changed: 28 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -60,7 +60,7 @@
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**方法一:记忆化搜索**
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63-
我们观察题目的数据范围,可以发现,`steps` 最大不超过 500ドル,ドル这意味着我们最远只能往右走 500ドル$ 步。
63+
我们观察题目的数据范围,可以发现 $steps$ 最大不超过 500ドル,ドル这意味着我们最远只能往右走 500ドル$ 步。
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6565
我们可以设计一个函数 $dfs(i, j),ドル表示当前在位置 $i,ドル并且剩余步数为 $j$ 的方案数。那么答案就是 $dfs(0, steps)$。
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@@ -72,7 +72,7 @@
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过程中,我们可以使用记忆化搜索避免重复计算。
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75-
时间复杂度 $O(steps \times steps),空间复杂度 O(steps \times steps)$。其中 $steps$ 是题目给定的步数。
75+
时间复杂度 $O(steps \times steps)$,空间复杂度 $O(steps \times steps)$。其中 $steps$ 是题目给定的步数。
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<!-- tabs:start -->
7878

@@ -198,6 +198,32 @@ func numWays(steps int, arrLen int) int {
198198
}
199199
```
200200

201+
### **TypeScript**
202+
203+
```ts
204+
function numWays(steps: number, arrLen: number): number {
205+
const f = Array.from({ length: steps }, () => Array(steps + 1).fill(-1));
206+
const mod = 10 ** 9 + 7;
207+
const dfs = (i: number, j: number) => {
208+
if (i > j || i >= arrLen || i < 0 || j < 0) {
209+
return 0;
210+
}
211+
if (i == 0 && j == 0) {
212+
return 1;
213+
}
214+
if (f[i][j] != -1) {
215+
return f[i][j];
216+
}
217+
let ans = 0;
218+
for (let k = -1; k <= 1; ++k) {
219+
ans = (ans + dfs(i + k, j - 1)) % mod;
220+
}
221+
return (f[i][j] = ans);
222+
};
223+
return dfs(0, steps);
224+
}
225+
```
226+
201227
### **...**
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203229
```

‎solution/1200-1299/1269.Number of Ways to Stay in the Same Place After Some Steps/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -168,6 +168,32 @@ func numWays(steps int, arrLen int) int {
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}
169169
```
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171+
### **TypeScript**
172+
173+
```ts
174+
function numWays(steps: number, arrLen: number): number {
175+
const f = Array.from({ length: steps }, () => Array(steps + 1).fill(-1));
176+
const mod = 10 ** 9 + 7;
177+
const dfs = (i: number, j: number) => {
178+
if (i > j || i >= arrLen || i < 0 || j < 0) {
179+
return 0;
180+
}
181+
if (i == 0 && j == 0) {
182+
return 1;
183+
}
184+
if (f[i][j] != -1) {
185+
return f[i][j];
186+
}
187+
let ans = 0;
188+
for (let k = -1; k <= 1; ++k) {
189+
ans = (ans + dfs(i + k, j - 1)) % mod;
190+
}
191+
return (f[i][j] = ans);
192+
};
193+
return dfs(0, steps);
194+
}
195+
```
196+
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### **...**
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```
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
1+
function numWays(steps: number, arrLen: number): number {
2+
const f = Array.from({ length: steps }, () => Array(steps + 1).fill(-1));
3+
const mod = 10 ** 9 + 7;
4+
const dfs = (i: number, j: number) => {
5+
if (i > j || i >= arrLen || i < 0 || j < 0) {
6+
return 0;
7+
}
8+
if (i == 0 && j == 0) {
9+
return 1;
10+
}
11+
if (f[i][j] != -1) {
12+
return f[i][j];
13+
}
14+
let ans = 0;
15+
for (let k = -1; k <= 1; ++k) {
16+
ans = (ans + dfs(i + k, j - 1)) % mod;
17+
}
18+
return (f[i][j] = ans);
19+
};
20+
return dfs(0, steps);
21+
}

‎solution/1200-1299/1275.Find Winner on a Tic Tac Toe Game/README.md‎

Lines changed: 98 additions & 37 deletions
Original file line numberDiff line numberDiff line change
@@ -83,11 +83,15 @@
8383

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<!-- 这里可写通用的实现逻辑 -->
8585

86-
判断 A、B 谁能获胜,只需判断最后一个落棋的人能否获胜即可。我们用数组 `counter` 记录 `0~2` 行、`0~2` 列、`正对角线`、`副对角线`是否已满 3 个棋子。如果等于 3,此人获胜,游戏结束。
86+
**方法一:判断最后一个落棋的人能否获胜**
8787

88-
若最后落棋者为未能获胜,棋盘被下满返回 `Draw`,未下满则返回 `Pending`。
88+
由于 `moves` 都有效,也即是说,不存在某个人获胜后,其他人仍然落棋的情况。因此,只需判断最后一个落棋的人能否获胜即可。
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90-
> 数组 `counter` 长度为 8,其中,`counter[0..2]` 对应 `0~2` 行,`counter[3..5]` 对应 `0~2` 列,`counter[6]` 对应正对角线,`counter[7]` 对应副对角线的落棋次数。
90+
我们用一个长度为 8ドル$ 的数组 `cnt` 记录行、列以及对角线的落棋次数。其中 $cnt[0, 1, 2]$ 分别表示第 0,ドル 1, 2$ 行的落棋次数,而 $cnt[3, 4, 5]$ 分别表示第 0,ドル 1, 2$ 列的落棋次数,另外 $cnt[6]$ 和 $cnt[7]$ 分别表示两条对角线的落棋次数。落棋过程中,如果某个人在某一行、列或对角线上落棋次数达到 3ドル$ 次,则该人获胜。
91+
92+
如果最后一个落棋的人没有获胜,那么我们判断棋盘是否已满,如果已满,则平局;否则,游戏尚未结束。
93+
94+
时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为 `moves` 的长度。
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9296
<!-- tabs:start -->
9397

@@ -99,22 +103,17 @@
99103
class Solution:
100104
def tictactoe(self, moves: List[List[int]]) -> str:
101105
n = len(moves)
102-
counter = [0] * 8
103-
for i in range(n - 1, -1, -2):
104-
row, col = moves[i][0], moves[i][1]
105-
counter[row] += 1
106-
counter[col + 3] += 1
107-
if row == col:
108-
counter[6] += 1
109-
if row + col == 2:
110-
counter[7] += 1
111-
if (
112-
counter[row] == 3
113-
or counter[col + 3] == 3
114-
or counter[6] == 3
115-
or counter[7] == 3
116-
):
117-
return "A" if (i % 2) == 0 else "B"
106+
cnt = [0] * 8
107+
for k in range(n - 1, -1, -2):
108+
i, j = moves[k]
109+
cnt[i] += 1
110+
cnt[j + 3] += 1
111+
if i == j:
112+
cnt[6] += 1
113+
if i + j == 2:
114+
cnt[7] += 1
115+
if any(v == 3 for v in cnt):
116+
return "B" if k & 1 else "A"
118117
return "Draw" if n == 9 else "Pending"
119118
```
120119

@@ -126,15 +125,19 @@ class Solution:
126125
class Solution {
127126
public String tictactoe(int[][] moves) {
128127
int n = moves.length;
129-
int[] counter = new int[8];
130-
for (int i = n - 1; i >= 0; i -= 2) {
131-
int row = moves[i][0], col = moves[i][1];
132-
++counter[row];
133-
++counter[col + 3];
134-
if (row == col) ++counter[6];
135-
if (row + col == 2) ++counter[7];
136-
if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
137-
return (i % 2) == 0 ? "A" : "B";
128+
int[] cnt = new int[8];
129+
for (int k = n - 1; k >= 0; k -= 2) {
130+
int i = moves[k][0], j = moves[k][1];
131+
cnt[i]++;
132+
cnt[j + 3]++;
133+
if (i == j) {
134+
cnt[6]++;
135+
}
136+
if (i + j == 2) {
137+
cnt[7]++;
138+
}
139+
if (cnt[i] == 3 || cnt[j + 3] == 3 || cnt[6] == 3 || cnt[7] == 3) {
140+
return k % 2 == 0 ? "A" : "B";
138141
}
139142
}
140143
return n == 9 ? "Draw" : "Pending";
@@ -149,22 +152,80 @@ class Solution {
149152
public:
150153
string tictactoe(vector<vector<int>>& moves) {
151154
int n = moves.size();
152-
vector<int> counter(8, 0);
153-
for (int i = n - 1; i >= 0; i -= 2) {
154-
int row = moves[i][0], col = moves[i][1];
155-
++counter[row];
156-
++counter[col + 3];
157-
if (row == col) ++counter[6];
158-
if (row + col == 2) ++counter[7];
159-
if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
160-
return (i % 2 == 0) ? "A" : "B";
155+
int cnt[8]{};
156+
for (int k = n - 1; k >= 0; k -= 2) {
157+
int i = moves[k][0], j = moves[k][1];
158+
cnt[i]++;
159+
cnt[j + 3]++;
160+
if (i == j) {
161+
cnt[6]++;
162+
}
163+
if (i + j == 2) {
164+
cnt[7]++;
165+
}
166+
if (cnt[i] == 3 || cnt[j + 3] == 3 || cnt[6] == 3 || cnt[7] == 3) {
167+
return k % 2 == 0 ? "A" : "B";
161168
}
162169
}
163170
return n == 9 ? "Draw" : "Pending";
164171
}
165172
};
166173
```
167174
175+
### **Go**
176+
177+
```go
178+
func tictactoe(moves [][]int) string {
179+
n := len(moves)
180+
cnt := [8]int{}
181+
for k := n - 1; k >= 0; k -= 2 {
182+
i, j := moves[k][0], moves[k][1]
183+
cnt[i]++
184+
cnt[j+3]++
185+
if i == j {
186+
cnt[6]++
187+
}
188+
if i+j == 2 {
189+
cnt[7]++
190+
}
191+
if cnt[i] == 3 || cnt[j+3] == 3 || cnt[6] == 3 || cnt[7] == 3 {
192+
if k%2 == 0 {
193+
return "A"
194+
}
195+
return "B"
196+
}
197+
}
198+
if n == 9 {
199+
return "Draw"
200+
}
201+
return "Pending"
202+
}
203+
```
204+
205+
### **TypeScript**
206+
207+
```ts
208+
function tictactoe(moves: number[][]): string {
209+
const n = moves.length;
210+
const cnt = new Array(8).fill(0);
211+
for (let k = n - 1; k >= 0; k -= 2) {
212+
const [i, j] = moves[k];
213+
cnt[i]++;
214+
cnt[j + 3]++;
215+
if (i == j) {
216+
cnt[6]++;
217+
}
218+
if (i + j == 2) {
219+
cnt[7]++;
220+
}
221+
if (cnt[i] == 3 || cnt[j + 3] == 3 || cnt[6] == 3 || cnt[7] == 3) {
222+
return k % 2 == 0 ? 'A' : 'B';
223+
}
224+
}
225+
return n == 9 ? 'Draw' : 'Pending';
226+
}
227+
```
228+
168229
### **...**
169230

170231
```

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