剑指offer--快速排序递归实现，年龄排序O(N)实现，旋转数组最小值查找

快速排序算法的递归实现：
void swap(int *a,int *b)
{
int c;
c=*a;
*a=*b;
*b=c;
}

int Partition(int data[],int length,int start,int end){
int i;
i= RandomInRange(start,end);
swap(&data[i],&data[end]);
int small;//计数左侧的数据个数
small = start-1;
for(i=start;i<end;i++)
{
if(data[i]<data[end])
{
small++;
if(small!=i)
swap(&data[i],&data[small]);
}

}
small++;
swap(&data[end],&data[small]);
return small;

}
void quickSort(int data[],int length,int start,int end)
{
if(start==end) return;

int index;
index = Partition(data,length,start,end);
if(index>start) quickSort(data,length,start,index-1);
if(index<end) quickSort(data,length,index+1,end);
}

对公司员工的年龄排序，复杂度为N。
void sortAges(int ares[],int length)
{
int i,j,index;
int numOfAges[100]={0};

if(ages==NULL) return;

for(i=0;i<length;i++)
{
if(ages[i]>99||ages<0) throw new std::exception();
numOfAges[ages[i]]++;
}
index = 0;
for(i=0;i<100;i++)
{
for(j=0;j<numOfAges[i];j++)
{
ages[index++]=i;
}

}
}

面试题八：旋转数组的最小数字。

<pre name="code" class="objc">int MinInOrder(int numbers[],int index1,int index2)
{
int result,i;
result = numbers[index1];
for(i=index1+1;i<=index2;i++)
{
if(result>numbers[index1])
result = numbers[index1];
}
return result;
}

int Min(int *numbers,int length)
{
int index1 = 0;
int index2 = length-1;
int indexMid=0;
if(numbers==NULL||length<=0)
{
throw new std::exception("invalid parameters.");
}
while(numbers[index1]>=numbers[index2])
{
if(index2-index1==1)
{
indexMid=  index2;
break;
}

indexMid=(index1+index2)/2;

if(numbers[index1]==numbers[index2]&&numbers[index2]==numbers[indexMid])
return MinInOrder(numbers,index1,index2);

if(numbers[index1]<=numbers[indexMid])
index1 = indexMid;
else if(numbers[index2]>=numbers[indexMid])
index2 = indexMid;
}

return numbers[indexMid];
}