我试图通过修改的二进制搜索在排序数组中定位旋转点。
考虑此数组int values[9]={7, 8, 9, 1, 2, 3, 4, 5, 6};
此处的旋转点位于索引= 3,即9。
我为上述操作编写了此函数。
void FindRotationPoint(int values[], int numvalues)
{
int first =0;
int last = numvalues-1;
int middle;
bool moreTosearch= (first<=last);
while(first<=last)
{
middle = (first+last)/2;
if(values[0]>values[middle]) //Keep looking right if the middle value in array is greater than first
{
last = middle-1;
}
if (values[0]<values[middle]) //Keep looking left if the middle value in array is less than first
{
first = middle+1;
}
}
cout<<middle+1<<endl;
cout<<values[middle];
}
如果元素是
int values[9]={7, 8, 9, 1, 2, 3, 4, 5, 6};
输出:4,1(错误)int values[9]={7, 8, 9, 10, 2, 3, 4, 5, 6};
输出:4,10(正确)发现在偶数位置的旋转点是正确的,而在其他情况下,它将找到后续元素。我在上面的代码中缺少什么?
最佳答案
这有效:
void FindRotationPoint(int values[], int numvalues)
{
int first =0;
int last = numvalues-1;
int middle=0;
bool moreTosearch= (first<=last);
while(first<last)
{
middle = (first+last)/2;
if(values[first]>=values[middle]) //Keep looking right if the middle value in array is greater than first
{
last = middle;
cout<<"first>middle: "<<first<<" "<<middle<<" "<<last<<"\n";
}
else if (values[middle]>=values[last]) //Keep looking left if the middle value in array is less than first
{
first = middle;
cout<<"middle<last: "<<first<<" "<<middle<<" "<<last<<"\n";
}
}
cout<<middle+1<<endl;
cout<<values[middle];
}
int main()
{
int values[9]={7, 8, 9, 1, 2, 3, 4, 5, 6};
FindRotationPoint(values, 9);
return 0;
}