6-9 Sort Three Distinct Keys （20 分）

最新推荐文章于 2023-09-27 12:16:52 发布

沿麓

最新推荐文章于 2023-09-27 12:16:52 发布

阅读量2.1k

点赞数 2

CC 4.0 BY-SA版权

分类专栏：数据结构 PTA

本文链接：https://blog.youkuaiyun.com/qq_40215528/article/details/84778047

数据结构同时被 2 个专栏收录

27 篇文章

订阅专栏

PTA

26 篇文章

订阅专栏

本文介绍了一个O(N)复杂度的算法，用于将包含true、false和maybe三种字符串的数组重新排列。该算法确保所有false元素位于maybe元素之前，而maybe元素又位于true元素之前，同时只使用常数级额外空间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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;
    }
}