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feat: add solutions to lc/lcci problems (doocs#1620)
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-74
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11 files changed

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-74
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‎lcci/08.01.Three Steps Problem/README.md

Lines changed: 20 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -113,6 +113,26 @@ class Solution:
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return sum(pow(a, n - 4)[0]) % mod
114114
```
115115

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```python
117+
import numpy as np
118+
119+
120+
class Solution:
121+
def waysToStep(self, n: int) -> int:
122+
if n < 4:
123+
return 2 ** (n - 1)
124+
mod = 10**9 + 7
125+
factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
126+
res = np.mat([(4, 2, 1)], np.dtype("O"))
127+
n -= 4
128+
while n:
129+
if n & 1:
130+
res = res * factor % mod
131+
factor = factor * factor % mod
132+
n >>= 1
133+
return res.sum() % mod
134+
```
135+
116136
### **Java**
117137

118138
<!-- 这里可写当前语言的特殊实现逻辑 -->

‎lcci/08.01.Three Steps Problem/README_EN.md

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@@ -75,6 +75,26 @@ class Solution:
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return sum(pow(a, n - 4)[0]) % mod
7676
```
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```python
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import numpy as np
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82+
class Solution:
83+
def waysToStep(self, n: int) -> int:
84+
if n < 4:
85+
return 2 ** (n - 1)
86+
mod = 10**9 + 7
87+
factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
88+
res = np.mat([(4, 2, 1)], np.dtype("O"))
89+
n -= 4
90+
while n:
91+
if n & 1:
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res = res * factor % mod
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factor = factor * factor % mod
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n >>= 1
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return res.sum() % mod
96+
```
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7898
### **Java**
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80100
```java

‎lcci/08.01.Three Steps Problem/Solution.py

Lines changed: 13 additions & 22 deletions
Original file line numberDiff line numberDiff line change
@@ -1,26 +1,17 @@
1-
class Solution:
2-
def waysToStep(self, n: int) -> int:
3-
mod = 10**9 + 7
4-
5-
def mul(a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
6-
m, n = len(a), len(b[0])
7-
c = [[0] * n for _ in range(m)]
8-
for i in range(m):
9-
for j in range(n):
10-
for k in range(len(a[0])):
11-
c[i][j] = (c[i][j] + a[i][k] * b[k][j] % mod) % mod
12-
return c
1+
import numpy as np
132

14-
def pow(a: List[List[int]], n: int) -> List[List[int]]:
15-
res = [[4, 2, 1]]
16-
while n:
17-
if n & 1:
18-
res = mul(res, a)
19-
n >>= 1
20-
a = mul(a, a)
21-
return res
223

4+
class Solution:
5+
def waysToStep(self, n: int) -> int:
236
if n < 4:
247
return 2 ** (n - 1)
25-
a = [[1, 1, 0], [1, 0, 1], [1, 0, 0]]
26-
return sum(pow(a, n - 4)[0]) % mod
8+
mod = 10**9 + 7
9+
factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
10+
res = np.mat([(4, 2, 1)], np.dtype("O"))
11+
n -= 4
12+
while n:
13+
if n & 1:
14+
res = res * factor % mod
15+
factor = factor * factor % mod
16+
n >>= 1
17+
return res.sum() % mod

‎solution/0000-0099/0070.Climbing Stairs/README.md

Lines changed: 19 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -60,7 +60,7 @@ $$
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。
6262

63-
**方法二：矩阵快速幂**
63+
**方法二：矩阵快速幂加速递推**
6464

6565
我们设 $Fib(n)$ 表示一个 $1 \times 2$ 的矩阵 $\begin{bmatrix} F_n & F_{n - 1} \end{bmatrix}$，其中 $F_n$ 和 $F_{n - 1}$ 分别是第 $n$ 个和第 $n - 1$ 个斐波那契数。
6666

@@ -128,7 +128,7 @@ class Solution:
128128
return c
129129

130130
def pow(a: List[List[int]], n: int) -> List[List[int]]:
131-
res = [[1, 1], [0, 0]]
131+
res = [[1, 1]]
132132
while n:
133133
if n & 1:
134134
res = mul(res, a)
@@ -140,6 +140,23 @@ class Solution:
140140
return pow(a, n - 1)[0][0]
141141
```
142142

143+
```python
144+
import numpy as np
145+
146+
147+
class Solution:
148+
def climbStairs(self, n: int) -> int:
149+
res = np.mat([(1, 1)], np.dtype("O"))
150+
factor = np.mat([(1, 1), (1, 0)], np.dtype("O"))
151+
n -= 1
152+
while n:
153+
if n & 1:
154+
res *= factor
155+
factor *= factor
156+
n >>= 1
157+
return res[0, 0]
158+
```
159+
143160
### **Java**
144161

145162
<!-- 这里可写当前语言的特殊实现逻辑 -->

‎solution/0000-0099/0070.Climbing Stairs/README_EN.md

Lines changed: 18 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -65,7 +65,7 @@ class Solution:
6565
return c
6666

6767
def pow(a: List[List[int]], n: int) -> List[List[int]]:
68-
res = [[1, 1], [0, 0]]
68+
res = [[1, 1]]
6969
while n:
7070
if n & 1:
7171
res = mul(res, a)
@@ -77,6 +77,23 @@ class Solution:
7777
return pow(a, n - 1)[0][0]
7878
```
7979

80+
```python
81+
import numpy as np
82+
83+
84+
class Solution:
85+
def climbStairs(self, n: int) -> int:
86+
res = np.mat([(1, 1)], np.dtype("O"))
87+
factor = np.mat([(1, 1), (1, 0)], np.dtype("O"))
88+
n -= 1
89+
while n:
90+
if n & 1:
91+
res *= factor
92+
factor *= factor
93+
n >>= 1
94+
return res[0, 0]
95+
```
96+
8097
### **Java**
8198

8299
```java

‎solution/1100-1199/1137.N-th Tribonacci Number/README.md

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@@ -122,6 +122,27 @@ class Solution:
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return sum(pow(a, n - 3)[0])
123123
```
124124

125+
```python
126+
import numpy as np
127+
128+
129+
class Solution:
130+
def tribonacci(self, n: int) -> int:
131+
if n == 0:
132+
return 0
133+
if n < 3:
134+
return 1
135+
factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
136+
res = np.mat([(1, 1, 0)], np.dtype("O"))
137+
n -= 3
138+
while n:
139+
if n & 1:
140+
res *= factor
141+
factor *= factor
142+
n >>= 1
143+
return res.sum()
144+
```
145+
125146
### **Java**
126147

127148
<!-- 这里可写当前语言的特殊实现逻辑 -->

‎solution/1100-1199/1137.N-th Tribonacci Number/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -80,6 +80,27 @@ class Solution:
8080
return sum(pow(a, n - 3)[0])
8181
```
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83+
```python
84+
import numpy as np
85+
86+
87+
class Solution:
88+
def tribonacci(self, n: int) -> int:
89+
if n == 0:
90+
return 0
91+
if n < 3:
92+
return 1
93+
factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
94+
res = np.mat([(1, 1, 0)], np.dtype("O"))
95+
n -= 3
96+
while n:
97+
if n & 1:
98+
res *= factor
99+
factor *= factor
100+
n >>= 1
101+
return res.sum()
102+
```
103+
83104
### **Java**
84105

85106
```java

‎solution/1100-1199/1137.N-th Tribonacci Number/Solution.py

Lines changed: 12 additions & 20 deletions
Original file line numberDiff line numberDiff line change
@@ -1,26 +1,18 @@
1-
class Solution:
2-
def tribonacci(self, n: int) -> int:
3-
def mul(a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
4-
m, n = len(a), len(b[0])
5-
c = [[0] * n for _ in range(m)]
6-
for i in range(m):
7-
for j in range(n):
8-
for k in range(len(a[0])):
9-
c[i][j] = c[i][j] + a[i][k] * b[k][j]
10-
return c
1+
import numpy as np
112

12-
def pow(a: List[List[int]], n: int) -> List[List[int]]:
13-
res = [[1, 1, 0]]
14-
while n:
15-
if n & 1:
16-
res = mul(res, a)
17-
n >>= 1
18-
a = mul(a, a)
19-
return res
203

4+
class Solution:
5+
def tribonacci(self, n: int) -> int:
216
if n == 0:
227
return 0
238
if n < 3:
249
return 1
25-
a = [[1, 1, 0], [1, 0, 1], [1, 0, 0]]
26-
return sum(pow(a, n - 3)[0])
10+
factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
11+
res = np.mat([(1, 1, 0)], np.dtype("O"))
12+
n -= 3
13+
while n:
14+
if n & 1:
15+
res *= factor
16+
factor *= factor
17+
n >>= 1
18+
return res.sum()

‎solution/1200-1299/1220.Count Vowels Permutation/README.md

Lines changed: 28 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -91,7 +91,7 @@ $$
9191

9292
时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 是字符串的长度，而 $C$ 是元音字母的个数。本题中 $C=5$。
9393

94-
**方法二：矩阵快速幂**
94+
**方法二：矩阵快速幂加速递推**
9595

9696
时间复杂度 $O(C^3 \times \log n)$，空间复杂度 $O(C^2)$，其中 $C$ 是元音字母的个数，本题中 $C=5$。
9797

@@ -151,6 +151,33 @@ class Solution:
151151
return sum(map(sum, res)) % mod
152152
```
153153

154+
```python
155+
import numpy as np
156+
157+
158+
class Solution:
159+
def countVowelPermutation(self, n: int) -> int:
160+
mod = 10**9 + 7
161+
factor = np.mat(
162+
[
163+
(0, 1, 0, 0, 0),
164+
(1, 0, 1, 0, 0),
165+
(1, 1, 0, 1, 1),
166+
(0, 0, 1, 0, 1),
167+
(1, 0, 0, 0, 0),
168+
],
169+
np.dtype("O"),
170+
)
171+
res = np.mat([(1, 1, 1, 1, 1)], np.dtype("O"))
172+
n -= 1
173+
while n:
174+
if n & 1:
175+
res = res * factor % mod
176+
factor = factor * factor % mod
177+
n >>= 1
178+
return res.sum() % mod
179+
```
180+
154181
### **Java**
155182

156183
<!-- 这里可写当前语言的特殊实现逻辑 -->

‎solution/1200-1299/1220.Count Vowels Permutation/README_EN.md

Lines changed: 27 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -119,6 +119,33 @@ class Solution:
119119
return sum(map(sum, res)) % mod
120120
```
121121

122+
```python
123+
import numpy as np
124+
125+
126+
class Solution:
127+
def countVowelPermutation(self, n: int) -> int:
128+
mod = 10**9 + 7
129+
factor = np.mat(
130+
[
131+
(0, 1, 0, 0, 0),
132+
(1, 0, 1, 0, 0),
133+
(1, 1, 0, 1, 1),
134+
(0, 0, 1, 0, 1),
135+
(1, 0, 0, 0, 0),
136+
],
137+
np.dtype("O"),
138+
)
139+
res = np.mat([(1, 1, 1, 1, 1)], np.dtype("O"))
140+
n -= 1
141+
while n:
142+
if n & 1:
143+
res = res * factor % mod
144+
factor = factor * factor % mod
145+
n >>= 1
146+
return res.sum() % mod
147+
```
148+
122149
### **Java**
123150

124151
```java

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