毛里求斯国旗问题

Question

我已经为荷兰国旗问题找到了解决方案。

但这一次，我想尝试一些更困难的事情：毛里求斯国旗问题 - 4 种颜色，而不是 3 种。对于有效的算法有什么建议吗？

基本上，毛里求斯国旗问题的重点是如何根据毛里求斯国旗的颜色顺序（红、蓝、黄、绿）对给定的对列表进行排序。并且数字也必须按升序排序。

方案编程示例 输入：(

(R . 3) (G . 6) (Y . 1) (B . 2) (Y . 7) (G . 3) (R . 1) (B . 8) )

输出

：( (R.1) (R.3) (B.2) (B.8) (Y.1) (Y.7) (G.3) (G.6) )

Answer 1

let a:string[] = ['1','2','1','0','2','4','3','0','1','3'];

function sort3(a:string[]):void{
    let low = 0;
    let mid1 = 0;
    let mid2 = 0;
    let mid3 = 0;
    let high = a.length - 1;
    while(mid3<=high){
        switch(a[mid3]){
            case '0': [a[mid3],a[low]] = [a[low],a[mid3]];
                    low++;
                    if(mid1 < low)
                    mid1++;
                    if(mid2 < mid1)
                    mid2++;
                    if(mid3 < mid2)
                    mid3++;
                    break;

            case '1': [a[mid3],a[mid1]] = [a[mid1],a[mid3]];
                    mid1++;
                    if(mid2 < mid1)
                    mid2++;
                    if(mid3 < mid2)
                    mid3++
                    break;

            case '2': [a[mid2],a[mid3]] = [a[mid3],a[mid2]];
                        mid2++;
                        mid3++;
                   break;

            case '3':
                    mid3++;break;

            case '4': [a[mid3],a[high]] = [a[high],a[mid3]];
                        high--;
        }
    }
}

Answer 2

这是我想出的。我使用数字而不是颜色。

// l  - index at which 0 should be inserted.
// m1 - index at which 1 should be inserted.
// m2 - index at which 2 should be inserted.
// h  - index at which 3 should be inserted.
l=m1=m2=0;
h=arr.length-1
while(m2 <= h) {
    if (arr[m2] == 0) {
        swap(arr, m2, l);
        l++;

        // m1 should be incremented if it is less than l as 1 can come after all
        // 0's
        //only.
        if (m1 < l) {
            m1++;
        }
        // Now why not always incrementing m2 as we used to do in 3 flag partition
        // while comparing with 0? Let's take an example here. Suppose arr[l]=1
        // and arr[m2]=0. So we swap arr[l] with arr[m2] with and increment l.
        // Now arr[m2] is equal to 1. But if arr[m1] is equal to 2 then we should
        // swap arr[m1] with arr[m2]. That's  why arr[m2] needs to be processed
        // again for the sake of arr[m1]. In any case, it should not be less than
        // l, so incrmenting.
        if(m2<l) {
            m2++;
        }       
    }
    // From here it is exactly same as 3 flag.
    else if(arr[m2]==1) {
        swap(arr, m1, m2)
        m1++;
        m2++;           
    }
    else if(arr[m2] ==2){
        m2++;
    }
    else {
        swap(arr, m2, h);
        h--;
    }           
}


}

同样，我们可以写五个标志。

    l=m1=m2=m3=0;
    h= arr.length-1;
    while(m3 <= h) {
        if (arr[m3] == 0) {
            swap(arr, m3, l);
            l++;
            if (m1 < l) {
                m1++;
            }
            if(m2<l) {
                m2++;
            }
            if(m3<l) {
                m3++;
            }

        }
        else if(arr[m3]==1) {
            swap(arr, m1, m3);
            m1++;
            if(m2<m1) {
                m2++;
            }
            if(m3<m1) {
                m3++;
            }   

        }
        else if(arr[m3] ==2){
            swap(arr,m2,m3);
            m2++;
            m3++;
        }
        else if(arr[m3]==3) {
            m3++;
        }
        else {
            swap(arr, m3, h);
            h--;
        }   
    }

Answer 3

这就像荷兰国旗问题一样，但我们有四种颜色。本质上适用相同的策略。假设我们有（其中 ^ 代表正在扫描的点）。

  RRRRBBB???????????YYYYGGGG
         ^

我们扫描红色

1. ，然后我们将第一个蓝色与当前节点
2. 蓝色交换我们不做任何事情
3. 黄色我们与最后一个交换？
4. 绿色，我们将最后一个黄色与最后一个交换？那么当前节点与交换的 ?

所以我们需要比平时多跟踪或多一个指针。

我们需要跟踪第一个蓝色、第一个 ?、最后一个 ?、最后一个 Y

一般来说，相同的策略适用于任意数量的颜色，但需要越来越多的交换。

Answer 4

基本上，维护以下

a[0-p] => '0'
a[p-q] => '1'
a[q-r] => '2'
a[r-s] => traversing! 
a[s-length] => '3'          

代码：

        int p=-1,q=-1,r=0,s=a.length-1;
        while(r<=s){
            if(a[r]==0){
                exchange(a,p+1,r);
                p++;r++;
                if(q!=-1)
                    q++;
            } else if (a[r]==1){
                if(q==-1)
                    q=p;
                exchange(a,q+1,r);
                q++;r++;
            } else if(a[r]==2) {
                r++;
            } else {
                exchange(a,r,s);
                s--;
            }

        }

Answer 5

我确实有类似的代码，但插入了

function sort(a:string[]){    
         let low = 0;
         let mid1 = 0;
         let mid2 = a.length-1;
         let high = a.length-1;

        while(mid1 <= mid2){
             switch(a[mid1]){

                 case '0':

                        [a[mid1],a[low]] = [a[low],a[mid1]];
                        mid1++;
                        low++;
                        break;

                 case '1':mid1++;break;

                 case '2':
                 case '3':[a[mid1],a[mid2]] = [a[mid2],a[mid1]];
                            mid2--;
                            break;

                 }
         }
    //sort 2 and 3
         while(mid1 <= high){
          switch(a[mid1]){
                 case '2': mid1++; break;
                 case '3': [a[mid1],a[high]] = [a[high],a[mid1]]
                            high--;
                            break;

           }
        }

}

Answer 6

function sort3(a:string[]):void{
    let low = 0;
    let mid1 = 0;
    let mid2 = 0;
    let high = a.length - 1;
    while(mid2<=high){
        switch(a[mid2]){
            case '0': [a[mid2],a[low]] = [a[low],a[mid2]];
                    low++;
                    if(mid1<low)
                    mid1++;
                    if(mid2<mid1)
                    mid2++;
                    break;

            case '1': [a[mid2],a[mid1]] = [a[mid1],a[mid2]];
                    mid1++;
                    mid2++;
                    break;

            case '2':mid2++
                   break;

            case '3':[a[mid2],a[high]] = [a[high],a[mid2]];
                    high--;
        }
    }
}