升序数组中寻找A[x] == x 的下标

朴素想法：

顺序遍历    复杂度 O（n);

int FoundIndex(int data[], int length){    for(int i = 0 ; i < length; ++i)    {        if(data[i] == i)            return i;    }    return -1;}

利用升序的性质  二分查找   复杂度 O（log(n));

int FoundIndex(int data[], int length, int start, int end){    if(data== NULL || start < 0|| end > length)    {        return -1;    }    int mid = (start + end) /2;    if( data[mid] == mid)    {        return mid;    }    else if(data[mid] > mid)    {        end = mid-1;       return  FoundIndex(data,length, start, end);    }    else if(data[mid] < mid)    {        start = mid+1;        return FoundIndex(data,length, start, end);    }}

int main(){    int data1[]= {2,2,2,4,6,7,7};   int  length = sizeof(data1)/sizeof(data[0]);   cout<<FoundIndex(data1,length)<<endl;    cout<<FoundIndex(data1,length, 0,length -1)<<endl;    return 0;}