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leetcode/problems/src/array/SortColors.java at master · KindleBooks66/leetcode · GitHub
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array
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SortColors.java
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problems
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src
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array
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SortColors.java
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package
array
;
/**
* Created by gouthamvidyapradhan on 06/08/2017. Given an array with n objects colored red, white or
* blue, sort them so that objects of the same color are adjacent, with the colors in the order red,
* white and blue.
*
* <p>Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue
* respectively.
*
* <p>Note: You are not suppose to use the library's sort function for this problem.
*
* <p>Follow up: A rather straight forward solution is a two-pass algorithm using counting sort.
* First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total
* number of 0's, then 1's and followed by 2's.
*
* <p>Could you come up with an one-pass algorithm using only constant space?
*
* <p>Solution: The below solution works with one pass. The basic idea is to keep track of start and
* end index of contiguous 1s and push the 0s to left of 1s and 2 to right of 1s.
*/
public
class
SortColors
{
/**
* Main method
*
* @param args
* @throws Exception
*/
public
static
void
main
(
String
[]
args
)
throws
Exception
{
int
[]
nums
= {
2
,
1
,
0
,
0
,
1
};
new
SortColors
().
sortColors
(
nums
);
for
(
int
i
:
nums
)
System
.
out
.
println
(
i
);
}
public
void
sortColors
(
int
[]
nums
) {
int
s
=
nums
[
0
];
// save the first index value
nums
[
0
] =
1
;
// overwrite with 1
int
l
=
0
,
r
=
0
;
// left and right index indicating the start and end index of 1s
for
(
int
i
=
1
;
i
<
nums
.
length
;
i
++) {
switch
(
nums
[
i
]) {
case
0
:
nums
[
l
] =
0
;
nums
[
r
+
1
] =
1
;
if
(
r
+
1
!=
i
) {
nums
[
i
] =
2
;
}
l
++;
r
++;
break
;
case
1
:
nums
[
r
+
1
] =
1
;
if
(
r
+
1
!=
i
) {
nums
[
i
] =
2
;
}
r
++;
break
;
}
}
// replace the initial overwritten value with the original value
if
(
s
==
0
)
nums
[
l
] =
0
;
else
if
(
s
==
2
)
nums
[
r
] =
2
;
}
}
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