根据给定索引对数组重新排序 [英] Reorder array according to given index

问题描述

算法根据给定索引对数组重新排序

Algorithm reorder array according to given index

a[] = [50, 40, 70, 60, 90]
 index[] = [3,  0,  4,  1,  2]
a= [60,50,90,40,70] 

在 O(n) 中并且没有额外的数组/空格

in O(n) and With out extra array/spaces

推荐答案

您将需要用于临时变量和循环计数器/索引的空间.根据算法通常的重新排序"也会将 index[] 改回 {0, 1, 2, 3, 4}.

You'll need space for a temp variable and loop counters / indices. The usual "reorder" according to algorithm is also going to change index[] back to {0, 1, 2, 3, 4}.

提示，注意 index[] 中索引的顺序.

Hint, noting the ordering of indices in index[].

          {0, 1, 2, 3, 4}
index[] = {3, 0, 4, 1, 2}

可以按照循环"进行重新排序.从 index[0] 开始，如果你查看 index[0]，然后查看 index[index[0]]，请注意循环"，依此类推...

The reordering can be done by following the "cycles". Start with index[0], and note the "cycles" if you look at index[0], then index[index[0]], and so on ...

// 1st cycle
index[0] == 3   // cycle starts at 0
index[3] == 1
index[1] == 0   // end of cycle since back at 0

// 2nd cycle
index[2] == 4   // cycle starts at 2
index[4] == 2   // end of cycle since back at 2

示例 C 代码:

#include <stdio.h>

static int A[] = {50, 40, 70, 60, 90};
static int I[]  = {3,  0,  4,  1,  2};

int main()
{
int i, j, k;
int tA;
    /* reorder A according to I */
    /* every move puts an element into place */
    /* time complexity is O(n) */
    for(i = 0; i < sizeof(A)/sizeof(A[0]); i++){
        if(i != I[i]){
            tA = A[i];
            j = i;
            while(i != (k = I[j])){
                A[j] = A[k];
                I[j] = j;
                j = k;
            }
            A[j] = tA;
            I[j] = j;
        }
    }
    for(i = 0; i < sizeof(A)/sizeof(A[0]); i++)
        printf("%d
", A[i]);
    return 0;
}

相同的算法，但使用交换而不是移动(这是较慢的方法).

The same algorithm, but using swaps instead of moves (this is slower method).

#include <stdio.h>

#define swap(a, b) {(a)^=(b); (b)^=(a); (a)^=(b);}

static int A[] = {50, 40, 70, 60, 90};
static int I[]  = {3,  0,  4,  1,  2};

int main()
{
int i, j, k;
    /* reorder A according to I */
    /* every swap puts an element into place */
    /* last swap of a cycle puts both elements into place */
    /* time complexity is O(n) */
    for(i = 0; i < sizeof(A)/sizeof(A[0]); i++){
        if(i != I[i]){
            j = i;
            while(i != (k = I[j])){
                swap(A[j], A[k]);
                I[j] = j;
                j = k;
            }
            I[j] = j;
        }
    }
    for(i = 0; i < sizeof(A)/sizeof(A[0]); i++)
        printf("%d
", A[i]);
    return 0;
}

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