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leetcode/algorithms/java/src/powXn/PowXn.java at master · wcoder0/leetcode · GitHub
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PowXn.java
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PowXn.java
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// Source : https://leetcode.com/problems/powx-n/
// Author : Tianming Cao
// Date : 2018-02-11
/**********************************************************************************
*
* Implement pow(x, n).
*
* Example 1:
*
* Input: 2.00000, 10
* Output: 1024.00000
*
* Example 2:
*
* Input: 2.10000, 3
* Output: 9.26100
*
**********************************************************************************/
package
powXn
;
public
class
PowXn
{
/**
* Divide-and-Conquer method
* For example:
*
* 3^9=(3^4)^2*3
* ↓
* 3^4=(3^2)^2
* ↓
* 3^2=3*3
* ↓
* 3=3
*
* So, both Space and Time are O(logN)
*/
public
double
recursion
(
double
x
,
long
n
) {
if
(
n
==
1
) {
return
x
;
}
//We'd better use unsigned right shift
double
half
=
recursion
(
x
,
n
>>>
1
);
if
((
n
&
1
) ==
0
) {
return
half
*
half
;
}
else
{
return
half
*
half
*
x
;
}
}
public
double
myPow01
(
double
x
,
int
n
) {
if
(
n
==
0
||
x
==
1
) {
return
1
;
}
// Avoid being out of bounds, we should cast int to long
long
m
=
n
;
double
result
=
recursion
(
x
,
Math
.
abs
(
m
));
if
(
n
>
0
) {
return
result
;
}
else
{
return
1
/
result
;
}
}
public
double
myPow02
(
double
x
,
int
n
) {
if
(
n
==
0
||
x
==
1
) {
return
1
;
}
// Avoid being out of bounds, we should cast int to long
long
m
=
n
;
double
result
=
bitFunction
(
x
,
Math
.
abs
(
m
));
if
(
n
>
0
) {
return
result
;
}
else
{
return
1
/
result
;
}
}
/**
* Solution with bit-manipulation
* For example:
* 9=1001
* 3^9=(3^1)^1*(3^2)^0*(3^4)^0*(3^8)^1
* Space is O(1), Time is O(logN)
*/
public
double
bitFunction
(
double
x
,
long
n
) {
double
multy
=
1
;
double
base
=
x
;
for
(
long
k
=
n
;
k
>=
1
;
k
>>>=
1
) {
if
((
k
&
1
) >
0
) {
multy
=
multy
*
base
;
}
base
*=
base
;
}
return
multy
;
}
}
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