LeetCode OJ
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.
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.
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 an one-pass algorithm using only constant space?
【最简单：计数排序】
class Solution {
public:
void sortColors(int A[], int n) {
int i = 0;
int j = 0;
int k = 0;
for(int p = 0; p < n; p ++)
{
if(A[p] == 0)
{
i ++;
}
else if(A[p] == 1)
{
j ++;
}
else
k ++;
}
for(int p = 0; p < n; p ++)
{
if(p < i)
A[p] = 0;
else if(p >= i && p < i + j)
A[p] = 1;
else
A[p] = 2;
}
}
};
【最直观：平移插入】
class Solution {
public:
void sortColors(int A[], int n) {
int i = -1;
int j = -1;
int k = -1;
for(int p = 0; p < n; p ++)
{
//根据第i个数字，挪动0~i-1串。
if(A[p] == 0)
{
A[++k] = 2; //2往后挪
A[++j] = 1; //1往后挪
A[++i] = 0; //0往后挪
}
else if(A[p] == 1)
{
A[++k] = 2;
A[++j] = 1;
}
else
A[++k] = 2;
}
}
};
【最暴力：分情况讨论】
class Solution
{
public:
void sortColors(int A[], int n)
{
int index0 = 0; //next position of 0
int index2 = n-1; //next position of 2while(A[index0] == 0)
index0 ++;
while(A[index2] == 2)
index2 --;for(int i = index0; i <= index2; i ++)
{
if(A[i] == 0 || A[i] == 2)
{
if(A[i] == 2 && A[index0] == 1 && A[index2] == 0)
{
int temp = A[index2];
A[index2] = A[i];
A[i] = temp;while(A[index2] == 2)
index2 --;temp = A[index0];
A[index0] = A[i];
A[i] = temp;while(A[index0] == 0)
index0 ++;if(index0 > i)
i = index0;
}
else if(A[i] == 2 && A[index0] == 2 && A[index2] == 0)
{
int temp = A[index2];
A[index2] = A[i];
A[i] = temp;while(A[index2] == 2)
index2 --;temp = A[index0];
A[index0] = A[i];
A[i] = temp;while(A[index0] == 0)
index0 ++;if(index0 > i)
i = index0;while(A[index2] == 2)
index2 --;
}
else if(A[i] == 0 && A[index0] == 2 && A[index2] == 1)
{
int temp = A[index0];
A[index0] = A[i];
A[i] = temp;while(A[index0] == 0)
index0 ++;temp = A[index2];
A[index2] = A[i];
A[i] = temp;while(A[index2] == 2)
index2 --;
}
else if(A[i] == 0 && A[index0] == 2 && A[index2] == 0)
{
int temp = A[index0];
A[index0] = A[i];
A[i] = temp;while(A[index0] == 0)
index0 ++;temp = A[index2];
A[index2] = A[i];
A[i] = temp;while(A[index2] == 2)
index2 --;while(A[index0] == 0)
index0 ++;if(index0 > i)
i = index0;
}
else
{
if(A[i] == 0)
{
int temp = A[index0];
A[index0] = A[i];
A[i] = temp;while(A[index0] == 0)
index0 ++;if(index0 > i)
i = index0;
}
else
{// A[i] == 2
int temp = A[index2];
A[index2] = A[i];
A[i] = temp;while(A[index2] == 2)
index2 --;
}
}
}
}
}
};
【LeetCode】Sort Colors,布布扣,bubuko.com