基数排序:降序 [英] Radix Sort: Descending

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本文介绍了基数排序:降序的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

这是我的RadixSort函数(升序):

Here is my RadixSort function (ascending):

void RadixSort (int a[], int n)
{
    int i, m=0, exp=1, b[MAX];
    for (i=0; i<n; i++)
    {
        if (a[i]>m)
            m=a[i];
    }
    while (m/exp>0)
    {
        int bucket[10]={0};
        for (i=0; i<n; i++)
            bucket[a[i]/exp%10]++;
        for (i=1; i<10; i++)
            bucket[i]+=bucket[i-1];
        for (i=n-1; i>=0; i--)
            b[--bucket[a[i]/exp%10]]=a[i];
        for (i=0; i<n;i++){
            a[i]=b[i];
        }
        exp*=10;
    }
}

我尝试通过替换将其更改为降序排列

I'm try to change this to a descending sort by replacing

for (i=0; i<n;i++) {
    a[i]=b[i];
}

使用

for (i=0; i<n;i++) {
    a[i]=b[n-i-1];
}

但是没有用. 我尝试使用:[705、1725、99、9170、7013]

But it didn't work. I tried with: [705, 1725, 99, 9170, 7013]

但是结果是:[9170,7013,1725,99,705]

But the result is: [9170, 7013, 1725, 99, 705]

最后一个值总是错误的.我有什么想念的吗?

The last value is always wrong. Is there anything I missed?

推荐答案

该问题试图在每次通过时反转数组,因为基数排序将顺序保留为相等值.第三遍之后,0705在0099之前结束(7> 0).在最后一遍,最高有效数字为0,因此保留了顺序,因此b [0] = 0705,b [1] = 0099,然后取反为a [] = {...,0099,0705}.

The issue is trying to reverse the array on each pass, since radix sort preserves the order on equal values. After the third pass, 0705 ends up before 0099 (7 > 0). On the last pass, the most significant digits are 0, so the order kept, so b[0] = 0705, b[1] = 0099, then gets reversed to a[] = {..., 0099, 0705}.

与其在每次通过后都进行反转,不如使用9位数字反转用于存储区的索引.对更改进行评论:

Instead of reversing after each pass, reverse the indexes used for bucket by using 9 - digit. The changes are commented:

void RadixSort (int a[], int n){
int i, m=0, exp=1, b[MAX];
    for (i=0; i<n; i++)
        if (a[i]>m)
            m=a[i];
    while (m/exp>0)
    {
        int bucket[10]={0};
        for (i=0; i<n; i++)
            bucket[9-a[i]/exp%10]++;         // changed this line
        for (i=1; i<10; i++)
            bucket[i]+=bucket[i-1];
        for (i=n-1; i>=0; i--)
            b[--bucket[9-a[i]/exp%10]]=a[i]; // changed this line
        for (i=0; i<n;i++){
            a[i]=b[i];                       // changed this line
        }
        exp*=10;
    }
}

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