传送门
nowcoder
leetcode
题目描述
实现函数 double Power(double base, int exponent)，求 base 的 exponent 次方。
不得使用库函数，同时不需要考虑大数问题。
C++ 代码 - nowcoder
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
|
class Solution {
public:
double Power(double base, int exp) {
if (exp == 0) return 1.0;
if (base == 0) return 0.0;
bool isNegative = false;
if (exp < 0) {
exp = -exp;
isNegative = true;
}
double res = Power(base * base, exp / 2);
if ((exp & 1) == 1) {
res *= base;
}
return isNegative ? 1 / res : res;
}
};
double Power(double base, int exponent) {
if (exponent == 0) return 1.0;
if (base == 0.0) return 0.0;
bool isNegative = false;
if (exponent < 0) {
exponent *= -1;
isNegative = true;
}
double res = base;
for (int i = 2; i <= exponent; i ++) {
res *= base;
}
return isNegative ? 1 / res : res;
}
class Solution {
public:
double Power(double base, int exponent) {
if (exponent == 0) return 1.0;
if (base == 0.0) return 0.0;
long exp = exponent;
if (exponent < 0) {
exp = exponent * (-1.0);
}
double res = 1.0;
while (exp != 0) {
if ((exp & 1) == 1) {
res *= base;
}
base *= base;
exp >>= 1;
}
return exponent < 0 ? 1 / res : res;
}
};
|
C++ 代码 - leetcode
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
|
class Solution {
public:
double myPow(double x, int n) {
if (n == 0) return 1.0;
long exp = n;
bool isNegative = false;
if (n < 0) {
exp = n * (-1.0);
isNegative = true;
}
double res = myPow(x * x, exp / 2);
if ((exp & 1) == 1) {
res *= x;
}
return isNegative ? 1 / res : res;
}
};
class Solution {
public:
double myPow(double x, int n) {
if (n == 0) return 1.0;
if (x == 0.0) return n;
long exp = n;
if (n < 0) {
exp = n * (-1.0);
}
double res = 1.0;
while (exp) {
if ((exp & 1) == 1) {
res *= x;
}
x *= x;
exp >>= 1;
}
return n < 0 ? 1 / res : res;
}
};
|