题目:有两个有序的数组A1和A2,内存在A1的末尾有足够多的空余空间容纳A2。请实现一个函数,把A2中的所有数字插入到A1中并且所有的数字是排序的。
解法:从后向前依次比较处理,减少移动次数,提高效率。
void SortA1A2(int A1[],int length1,int sizeofA1,int A2[],int length2)
{
if(A1==NULL||A2==NULL||length1<=0||length2<=0||sizeofA1<=0)
return;
if(sizeofA1<length1+length2)
return;
int newlength1=length1+length2-1;
int plen1=length1-1;
int plen2=length2-1;
while (plen1>=0&&plen2>=0)
{
if(A1[plen1]>=A2[plen2])
A1[newlength1--]=A1[plen1--];
else
A1[newlength1--]=A2[plen2--];
}
while (plen2>=0)
{
A1[newlength1--]=A2[plen2--];
}
return;
}
int _tmain(int argc, _TCHAR* argv[])
{
const int length = 100;
int length1=5;
int length2=5;
int A1[length] = {1,3,5,7,9};
int A2[] = {2,4,6,8,10};
int A3[] ={12,14,16,18,20};
for(int i=0;i<length1;i++)
{
cout<<A1[i]<<" ";
}
for(int i=0;i<length2;i++)
{
cout<<A2[i]<<" ";
}
SortA1A2(A1,length1,length,A2,length2);
for(int i=0;i<length1+length2;i++)
{
cout<<A1[i]<<" ";
}
return 0;
}
参考:剑指offer 面试题4后扩展