75. Sort Colors

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.

click to show follow up.

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?

这道题有四种解法，个人认为第二种最为巧妙，3 pointers，用这种方法，不管给几种颜色都能应付。
// two pass O(m+n) space
void sortColors(int A[], int n) {
int num0 = 0, num1 = 0, num2 = 0;

for(int i = 0; i < n; i++) {
if (A[i] == 0) ++num0;
else if (A[i] == 1) ++num1;
else if (A[i] == 2) ++num2;
}

for(int i = 0; i < num0; ++i) A[i] = 0;
for(int i = 0; i < num1; ++i) A[num0+i] = 1;
for(int i = 0; i < num2; ++i) A[num0+num1+i] = 2;
}

// one pass in place solution
void sortColors(int A[], int n) {
int n0 = -1, n1 = -1, n2 = -1;
for (int i = 0; i < n; ++i) {
if (A[i] == 0)
{
A[++n2] = 2; A[++n1] = 1; A[++n0] = 0;
}
else if (A[i] == 1)
{
A[++n2] = 2; A[++n1] = 1;
}
else if (A[i] == 2)
{
A[++n2] = 2;
}
}
}

// one pass in place solution
void sortColors(int A[], int n) {
int j = 0, k = n - 1;
for (int i = 0; i <= k; ++i){
if (A[i] == 0 && i != j)
swap(A[i--], A[j++]);
else if (A[i] == 2 && i != k)
swap(A[i--], A[k--]);
}
}

// one pass in place solution
void sortColors(int A[], int n) {
int j = 0, k = n-1;
for (int i=0; i <= k; i++) {
if (A[i] == 0)
swap(A[i], A[j++]);
else if (A[i] == 2)
swap(A[i--], A[k--]);
}
}