LeetCode-Sort Colors

LeetCode-Sort Colors

題目信息：

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?

解題思路：使用兩個指針pointA代表0的位置，pointB代表2的位置。用一個for循環進行搜索，（1）遇到0的情況，直接存儲到pointA的位置。（2）遇到2的情況，要進行判斷，1）vpointB的位置是0嗎，2）pointB位置是2嗎，3）pointB的位置正好等於搜索的位置，對這三個情況進行處理。最後再用for[pointA,pointB]賦值為1就行。

class Solution {
public:
    void sortColors(int A[], int n) {
        int pointA = 0;
        int pointB = n-1;
        for (int i = 0; i <= pointB; i++) {
            if (A[i] == 0)
                A[pointA++] = 0;
            else if (A[i] == 2) {
                   if (A[pointB] == 0) {
                       A[pointA++] = 0 ;
                       A[pointB--] = 2;
                   } else if(A[pointB] == 2) {
                     if(i == pointB) {
                       pointB--;
                     } else {
                       while(A[pointB] == 2) {
                         pointB--;
                         if(i == pointB)
                         {
                           break;
                         }
                       }
                       if(A[pointB]== 0) {
                         A[pointA++] = 0 ;
                         A[pointB--] = 2;
                       } else {
                         A[pointB--] = 2;
                       }
                     }
                   } else A[pointB--] = 2;
            }
        }
        for(int i = pointA;i <= pointB;i++)
            A[i] = 1;
    }
};