【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
Sort Colors
Given an array with
n
objects colored red, white or blue, sort them
in-place
so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
Note:
You are not suppose to use the library's sort function for this problem.
Example:
Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]
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.
-
Could you come up with a one-pass algorithm using only constant space?
//问题：颜色排序（以后看可不可以用partition）
//双指针法，只能用于只有三类的排序，不过只需一次遍历
#include <algorithm>
class
Solution
{
public
:
void
sortColors
(
vector
<
int
>&
A
)
{
int
left
=
0
,
right
=
A
.
size
()-
1
;
//
双指针，
left
从左边扫描，
right
从右边扫描
for
(
int
i
=
0
;
i
<=
right
;
i
++)
//
扫描到
right
即可
{
if
(
A
[
i
]
==
0
)
swap
(
A
[
i
++],
A
[
left
++]);
//
将
0
换到左边
else
if
(
A
[
i
]
==
2
)
swap
(
A
[
i
--],
A
[
right
--]);
//
将
2
换到右边，
i--
是以免
right
换过来的值为
0（抵消for循环中的i++）
}
}
};
//存储后移法（相当于直接插入排序,只不过预先知道了有哪些数，故简单一点）
class
Solution
{
public
:
void
sortColors
(
vector
<
int
>&
A
)
{
int
n0
,
n1
,
n2
;
//类别索引
n0
=
n1
=
n2
=
0
;
for
(
int
i
=
0
;
i
<
A
.
size
();
i
++)
{
switch
(
A
[
i
])
{
case
0
:
A
[
n2
++]
=
2
;
A
[
n1
++]
=
1
;
A
[
n0
++]
=
0
;
//后面的元素往后移
break
;
case
1
:
A
[
n2
++]
=
2
;
A
[
n1
++]
=
1
;
break
;
case
2
:
A
[
n2
++]
=
2
;
}
}
}
};
//
计数排序，需要两次遍历
class
Solution
{
public
:
void
sortColors
(
vector
<
int
>&
nums
)
{
int
count
[
3
]
=
{
0
};
//
用于统计每个颜色出现的次数
for
(
int
i
=
0
;
i
<
nums
.
size
();
i
++)
//
统计直方图
count
[
nums
[
i
]]++;
for
(
int
i
=
0
,
index
=
0
;
i<3
;
i
++)
//
排序
for
(
int
j
=
0
;
j
<
count[i
];
j
++)
nums
[
index
++]
=
i
;
}
};