[LeetCode] Sort Colors 對於元素取值有限的數組，只遍歷一遍的排序方法

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?

 

我的解法是定義指針start = -1，指針end = n；一次遍歷，如果遇到0，交換給start，遇到2，交換給end，遇到1別管。

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;
    }
};

 

網上還有更加通用和直觀的解法

 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;
        }
    }

}

稍微簡化一點：

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;
       }
    }
}