6-9 Sort Three Distinct Keys （20 分）

Suppose you have an array of N elements, containing three distinct keys, "true", "false", and "maybe". Given an O(N) algorithm to rearrange the list so that all "false" elements precede "maybe" elements, which in turn precede "true" elements. You may use only constant extra space.

Format of functions:

void MySort( ElementType A[], int N );

where ElementType A[] contains the N elements.

Sample program of judge:

#include <stdio.h>
#include <stdlib.h>

typedef enum { true, false, maybe } Keys;
typedef Keys ElementType;

void Read( ElementType A[], int N ); /* details omitted */

void MySort( ElementType A[], int N );

void PrintA( ElementType A[], int N )
{
    int i, k;

    k = i = 0;
    for ( ; i<N && A[i]==false; i++ );
    if ( i > k )
        printf("false in A[%d]-A[%d]\n", k, i-1);
    k = i;
    for ( ; i<N && A[i]==maybe; i++ );
    if ( i > k )
        printf("maybe in A[%d]-A[%d]\n", k, i-1);
    k = i;
    for ( ; i<N && A[i]==true; i++ );
    if ( i > k )
        printf("true in A[%d]-A[%d]\n", k, i-1);
    if ( i < N )
        printf("Wrong Answer\n");
}

int main()
{
    int N;
    ElementType *A;

    scanf("%d", &N);
    A = (ElementType *)malloc(N * sizeof(ElementType));
    Read( A, N );
    MySort( A, N );
    PrintA( A, N );
    return 0;
}

/* Your function will be put here */

Sample Input:

6
2 2 0 1 0 0

Sample Output:

false in A[0]-A[0]
maybe in A[1]-A[2]
true in A[3]-A[5]

void MySort( ElementType A[], int N ){
    int a[3]={0};
    for (int i=0; i<N; i++) {
        switch (A[i]) {
            case 0:
                a[0]++;
                break;
            case 1:
                a[1]++;
                break;
            case 2:
                a[2]++;
                break;
            default:
                break;
        }
    }
    for (int i=0; i<a[1]; i++) {
        A[i]=1;
    }
    for (int i=a[1]; i<a[1]+a[2]; i++) {
        A[i]=2;
    }
    for (int i=a[1]+a[2]; i<a[1]+a[2]+a[0]; i++) {
        A[i]=0;
    }
}