[LeetCode] Sort Colors 对于元素取值有限的数组，只遍历一遍的排序方法

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?

 

我的解法是定义指针start = -1，指针end = n；一次遍历，如果遇到0，交换给start，遇到2，交换给end，遇到1别管。

class Solution {
public:
    void sortColors(int A[], int n) {
        if(n <= 1) return;
        int start = -1, end = n;
        int p = 0;
        while(p < n){
            if(A[p] == 0){
                if(p > start) swap(A, ++start, p);
                else ++p;
            }else if(A[p] == 2){
                if(p < end) swap(A, p, --end);
                else ++p;
            }else ++p;
        }
    }
private:
    void swap(int A[], int i, int j){
        int temp = A[i];
        A[i] = A[j];
        A[j] = temp;
    }
};

网上还有更加通用和直观的解法

 public void sortColors(int[] A) {


    int i=-1, j=-1, k=-1;

    for(int p = 0; p < A.length; p++)
    {
        if(A[p] == 0)
        {
            A[++k]=2;
            A[++j]=1;
            A[++i]=0;
        }
        else if (A[p] == 1)
        {
            A[++k]=2;
            A[++j]=1;

        }
        else if (A[p] == 2)
        {
            A[++k]=2;
        }
    }

}

稍微简化一点：

public void sortColors(int[] A) {

    int i=-1, j=-1;

    for(int p = 0; p < A.length; p++) {

       int v = A[p];
       A[p] = 2;

       if (v == 0) {

          A[++j] = 1;
          A[++i] = 0;
       }
       else if (v == 1) {

           A[++j] = 1;
       }
    }
}