如何将具有不同值的两个数组合并为一个数组? [英] How do I merge two arrays having different values into one array?

问题描述

假设您有一个数组 a[]=1,2,4,6 和第二个数组 b[]=3,5,7.合并结果应具有所有值，即 c[]=1,2,3,4,5,6,7.合并应该在不使用 中的函数的情况下完成.

Suppose you have one array a[]=1,2,4,6 and a second array b[]=3,5,7. The merged result should have all the values, i.e. c[]=1,2,3,4,5,6,7. The merge should be done without using functions from <string.h>.

推荐答案

我还没有编译和测试以下代码，但我相当有信心.我假设两个输入数组都已经排序.与仅针对此示例的解决方案相比，要实现此通用目的还有更多工作要做.毫无疑问，我确定的两个阶段可以合并，但可能更难阅读和验证；

I haven't compiled and tested the following code, but I am reasonably confident. I am assuming both input arrays are already sorted. There is more work to do to make this general purpose as opposed to a solution for this example only. No doubt the two phases I identify could be combined, but perhaps that would be harder to read and verify;

void merge_example()
{
    int a[] = {1,2,4,6};
    int b[] = {3,5,7};
    int c[100];     // fixme - production code would need a robust way
                    //  to ensure c[] always big enough
    int nbr_a = sizeof(a)/sizeof(a[0]);
    int nbr_b = sizeof(b)/sizeof(b[0]);
    int i=0, j=0, k=0;

    // Phase 1) 2 input arrays not exhausted
    while( i<nbr_a && j<nbr_b )
    {
        if( a[i] <= b[j] )
            c[k++] = a[i++];
        else
            c[k++] = b[j++];
    }

    // Phase 2) 1 input array not exhausted
    while( i < nbr_a )
        c[k++] = a[i++];
    while( j < nbr_b )
        c[k++] = b[j++];
}

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