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

快速排序的思想可以运用到很多情况，如，还可以用来实现，

1. 在长度为N的数组中，查找第K大的数字，

2.数组中出现次数超过一半的数字。

3.最小的K个数字。

等

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];}