Sort Colors--Leetcode

Given an array with n objects colored red, white or blue, sort them so, objects of the same color is Adjacen T, with the colors in the order red, white and blue.

Here, we'll use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note:
You is not a 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 a one-pass algorithm using only constant space?

Algorithm one: Red, white, and blue three hands moving backwards from the front

This algorithm understands that the brush wall can be borrowed.

Suppose there are x red, y white, z Blue

First, all blue (x+y+z).

Then brush (x+y) a white.

Then brush the x intenseness from scratch.

Class Solution {public:    void Sortcolors (int. a[], int n) {        int red =-1, white =-1, blue =-1;        for (int i=0; i<n; i++) {            if (a[i] = = 2) {                ++blue;            }            else if (a[i] = = 1) {                A[++blue] = 2;                A[++white] = 1;            }            else {                A[++blue] = 2;                A[++white] = 1;                A[++red] = 0;}}}    ;

Algorithm reference from:

Https://leetcode.com/discuss/1827/anyone-with-one-pass-and-constant-space-solution

Algorithm two: Red and white two hands in front of the back, blue hands moving forward from behind

Class Solution {public:    void Sortcolors (int. a[], int n) {        int red =-1, white = 0, blue = n;        while (White < blue) {            if (a[white] = = 1) {                ++white;            }            else if (a[white] = = 0) {                a[white++] = 1;                A[++red] = 0;            }            else {                A[white] = A[--blue];                A[blue] = 2;}}}    ;

Sort Colors--Leetcode