如何将具有不同值的两个数组合并为一个数组?

本文介绍了如何将具有不同值的两个数组合并为一个数组?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

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++]; }