Skip to content

Commit 90dbf6a

Browse files
yanglbmeyanglbme
committed
feat: add solutions to lc/lcof problems: Pow(x, n)
1 parent f055e23 commit 90dbf6a

File tree

17 files changed

+560
-216
lines changed

17 files changed

+560
-216
lines changed

‎lcof/面试题16. 数值的整数次方/README.md

Lines changed: 119 additions & 80 deletions
Original file line numberDiff line numberDiff line change
@@ -42,132 +42,171 @@
4242

4343
## 解法
4444

45+
<!-- 这里可写通用的实现逻辑 -->
46+
47+
**方法一：数学（快速幂）**
48+
49+
快速幂算法的核心思想是将幂指数 $n$ 拆分为若干个二进制位上的 $1$ 的和，然后将 $x$ 的 $n$ 次幂转化为 $x$ 的若干个幂的乘积。
50+
51+
时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 为幂指数。
52+
4553
<!-- tabs:start -->
4654

4755
### **Python3**
4856

57+
<!-- 这里可写当前语言的特殊实现逻辑 -->
58+
4959
```python
5060
class Solution:
5161
def myPow(self, x: float, n: int) -> float:
52-
if n == 0:
53-
return 1
54-
if n == 1:
55-
return x
56-
if n == -1:
57-
return 1 / x
58-
half = self.myPow(x, n // 2)
59-
return half * half * self.myPow(x, n % 2)
62+
def qmi(a, k):
63+
res = 1
64+
while k:
65+
if k & 1:
66+
res *= a
67+
a *= a
68+
k >>= 1
69+
return res
70+
71+
return qmi(x, n) if n >= 0 else 1 / qmi(x, -n)
6072
```
6173

6274
### **Java**
6375

76+
<!-- 这里可写当前语言的特殊实现逻辑 -->
77+
6478
```java
6579
class Solution {
6680
public double myPow(double x, int n) {
67-
if (n == 0) return 1;
68-
if (n == 1) return x;
69-
if (n == -1) return 1 / x;
70-
double half = myPow(x, n / 2);
71-
return half * half * myPow(x, n % 2);
81+
long N = n;
82+
return n >= 0 ? qmi(x, N) : 1.0 / qmi(x, -N);
83+
}
84+
85+
private double qmi(double a, long k) {
86+
double res = 1;
87+
while (k != 0) {
88+
if ((k & 1) != 0) {
89+
res *= a;
90+
}
91+
a *= a;
92+
k >>= 1;
93+
}
94+
return res;
7295
}
7396
}
7497
```
7598

76-
### **JavaScript**
99+
### **C++**
77100

78-
```js
79-
/**
80-
* @param {number} x
81-
* @param {number} n
82-
* @return {number}
83-
*/
84-
var myPow = function (x, n) {
85-
let r = 1;
86-
let tmp = x;
87-
let tag = 0;
88-
if (n < 0) {
89-
tag = 1;
90-
n = -n;
101+
```cpp
102+
class Solution {
103+
public:
104+
double myPow(double x, int n) {
105+
long long N = n;
106+
return N >= 0 ? qmi(x, N) : 1.0 / qmi(x, -N);
91107
}
92-
while (n) {
93-
if (n & 1) {
94-
r *= tmp;
108+
109+
double qmi(double a, long long k) {
110+
double res = 1;
111+
while (k) {
112+
if (k & 1) {
113+
res *= a;
114+
}
115+
a *= a;
116+
k >>= 1;
95117
}
96-
tmp *= tmp;
97-
n >>>= 1;
118+
return res;
98119
}
99-
return tag ? 1 / r : r;
100120
};
101121
```
102122

103123
### **Go**
104124

105125
```go
106126
func myPow(x float64, n int) float64 {
107-
p := abs(n)
108-
res := 1.0
109-
for p != 0 {
110-
if p&1 == 1 {
111-
res *= x
112-
}
113-
x *= x
114-
p = p >>1
115-
}
116-
if n < 0 {
117-
return 1/res
118-
}
119-
return res
127+
if n >= 0 {
128+
return qmi(x, n)
129+
}
130+
return 1.0 / qmi(x, -n)
120131
}
121132

122-
func abs(x int) int {
123-
if x > 0 {
124-
return x
125-
}
126-
return -x
133+
func qmi(a float64, k int) float64 {
134+
var res float64 = 1
135+
for k != 0 {
136+
if k&1 == 1 {
137+
res *= a
138+
}
139+
a *= a
140+
k >>= 1
141+
}
142+
return res
127143
}
128-
129144
```
130145

131-
### **C++**
146+
### **JavaScript**
132147

133-
```cpp
134-
class Solution {
135-
public:
136-
double myPow(double x, int n) {
137-
long m = n;
138-
if (m < 0) {
139-
x = 1 / x;
140-
m = -m;
148+
```js
149+
/**
150+
* @param {number} x
151+
* @param {number} n
152+
* @return {number}
153+
*/
154+
var myPow = function (x, n) {
155+
return n >= 0 ? qmi(x, n) : 1 / qmi(x, -n);
156+
};
157+
158+
function qmi(a, k) {
159+
let res = 1;
160+
while (k) {
161+
if (k & 1) {
162+
res *= a;
141163
}
142-
double ans = 1;
143-
while (m) {
144-
if (m & 1) {
145-
ans *= x;
146-
}
147-
x *= x;
148-
m >>= 1;
164+
a *= a;
165+
k >>>= 1;
166+
}
167+
return res;
168+
}
169+
```
170+
171+
### **TypeScript**
172+
173+
```ts
174+
function myPow(x: number, n: number): number {
175+
return n >= 0 ? qmi(x, n) : 1 / qmi(x, -n);
176+
}
177+
178+
function qmi(a: number, k: number): number {
179+
let res = 1;
180+
while (k) {
181+
if (k & 1) {
182+
res *= a;
149183
}
150-
return ans;
184+
a *= a;
185+
k >>>= 1;
151186
}
152-
};
187+
return res;
188+
}
153189
```
154190

155191
### **C#**
156192

157193
```cs
158194
public class Solution {
159195
public double MyPow(double x, int n) {
160-
if (n == 0) {
161-
return 1;
162-
}
163-
if (n == 1) {
164-
return x;
165-
}
166-
if (n == -1) {
167-
return 1 / x;
196+
long N = n;
197+
return n >= 0 ? qmi(x, N) : 1.0 / qmi(x, -N);
198+
}
199+
200+
private double qmi(double a, long k) {
201+
double res = 1;
202+
while (k != 0) {
203+
if ((k & 1) != 0) {
204+
res *= a;
205+
}
206+
a *= a;
207+
k >>= 1;
168208
}
169-
double half = MyPow(x, n / 2);
170-
return half * half * MyPow(x, n % 2);
209+
return res;
171210
}
172211
}
173212
```

‎lcof/面试题16. 数值的整数次方/Solution.cpp

Lines changed: 13 additions & 13 deletions
Original file line numberDiff line numberDiff line change
@@ -1,19 +1,19 @@
11
class Solution {
22
public:
33
double myPow(double x, int n) {
4-
long m = n;
5-
if (m < 0) {
6-
x = 1 / x;
7-
m = -m;
8-
}
9-
double ans = 1;
10-
while (m) {
11-
if (m & 1) {
12-
ans *= x;
4+
long long N = n;
5+
return N >= 0 ? qmi(x, N) : 1.0 / qmi(x, -N);
6+
}
7+
8+
double qmi(double a, long long k) {
9+
double res = 1;
10+
while (k) {
11+
if (k & 1) {
12+
res *= a;
1313
}
14-
x *= x;
15-
m >>= 1;
14+
a *= a;
15+
k >>= 1;
1616
}
17-
return ans;
17+
return res;
1818
}
19-
};
19+
};

‎lcof/面试题16. 数值的整数次方/Solution.cs

Lines changed: 13 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,18 @@
11
public class Solution {
22
public double MyPow(double x, int n) {
3-
if (n == 0) {
4-
return 1;
5-
}
6-
if (n == 1) {
7-
return x;
8-
}
9-
if (n == -1) {
10-
return 1 / x;
3+
long N = n;
4+
return n >= 0 ? qmi(x, N) : 1.0 / qmi(x, -N);
5+
}
6+
7+
private double qmi(double a, long k) {
8+
double res = 1;
9+
while (k != 0) {
10+
if ((k & 1) != 0) {
11+
res *= a;
12+
}
13+
a *= a;
14+
k >>= 1;
1115
}
12-
double half = MyPow(x, n / 2);
13-
return half * half * MyPow(x, n % 2);
16+
return res;
1417
}
1518
}

‎lcof/面试题16. 数值的整数次方/Solution.go

Lines changed: 13 additions & 18 deletions
Original file line numberDiff line numberDiff line change
@@ -1,23 +1,18 @@
11
func myPow(x float64, n int) float64 {
2-
p := abs(n)
3-
res := 1.0
4-
for p != 0 {
5-
if p&1 == 1 {
6-
res *= x
7-
}
8-
x *= x
9-
p = p >> 1
10-
}
11-
if n < 0 {
12-
return 1 / res
2+
if n >= 0 {
3+
return qmi(x, n)
134
}
14-
return res
5+
return 1.0 / qmi(x, -n)
156
}
167

17-
func abs(x int) int {
18-
if x > 0 {
19-
return x
8+
func qmi(a float64, k int) float64 {
9+
var res float64 = 1
10+
for k != 0 {
11+
if k&1 == 1 {
12+
res *= a
13+
}
14+
a *= a
15+
k >>= 1
2016
}
21-
return -x
22-
}
23-
17+
return res
18+
}

‎lcof/面试题16. 数值的整数次方/Solution.java

Lines changed: 14 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -1,9 +1,18 @@
11
class Solution {
22
public double myPow(double x, int n) {
3-
if (n == 0) return 1;
4-
if (n == 1) return x;
5-
if (n == -1) return 1 / x;
6-
double half = myPow(x, n / 2);
7-
return half * half * myPow(x, n % 2);
3+
long N = n;
4+
return n >= 0 ? qmi(x, N) : 1.0 / qmi(x, -N);
5+
}
6+
7+
private double qmi(double a, long k) {
8+
double res = 1;
9+
while (k != 0) {
10+
if ((k & 1) != 0) {
11+
res *= a;
12+
}
13+
a *= a;
14+
k >>= 1;
15+
}
16+
return res;
817
}
918
}

0 commit comments

Comments
 (0)