美团面试题：寻找数组置尾操作的最小值

#include <iostream>
using namespace std;
int getInvertNumber(int arr[],int nLength);


void main()
{
	int arr[] = {1,2,3,4,5,6,7,8,9};
	int brr[] = {1,3,4,6,8,9};
	int sb = getInvertNumber(brr,sizeof(brr)/sizeof(brr[0]));
	cout<<sb<<endl;


}


int getInvertNumber(int arr[],int nLength)
{
	int nLeft = 0;
	int nRight = nLength-1;
	int nMid;
	if(arr[nLeft]<arr[nRight])
		return 0;
	while(nLeft<nRight)
	{
		nMid = ( nLeft + nRight ) / 2;
		if(arr[nMid]>arr[nLeft])
			nLeft = nMid;
		else
			nRight = nMid;
	}


	return nLength -nMid-1;
}

美团面试题：寻找数组置尾操作的最小值

原文：http://blog.csdn.net/ddupd/article/details/40511943