21年考研王道数据结构2-9算法：通过折半查找指定值X并根据结果进行操作（两种方法）

using namespace std;

int ch(int a[], int left, int right, int x,int len)
{
    if (left > right) {
        cout"left>right! break;"endl;
            return -1;
    }
    //system("pause");
    cout "now is "2", a[i]= "2]", left= "      ", right= " ", x= " ", len=" endl;
    if (x == a[(left + right) / 2]&&((left + right) / 2+1 !=len)){ 
//找到X，排除掉X为最后一个元素的情况后开始和后一个元素交换位置
        cout "index=" 2 + 1  endl;
        int temp = a[(left + right) / 2];
        a[(left + right) / 2] = a[(left + right) / 2 + 1]; 
//不排除上述的情况话，会在此处出现内存错误，
//因为a[(left+right)/2]=a[9]=19,a[(left+right)/2+1]=a[10]未定义
        a[(left + right) / 2 + 1] = temp;
        return 1;

    }
    else if (x != a[(left + right) / 2] && left == right) { 
//未找到X，开始插入
        cout "not found!"  endl;
        for (int i = len-1;i > (left + right) / 2; i--) 
//这里i不能是len=10，因为同样a[10]无定义
        {
           a[i] = a[i - 1]; 
        } 
        a[(left + right) / 2] = x;
        return 0;
    }
    else if (x > a[(left + right) / 2]) return ch(a, (left + right) / 2+1, right, x,len);
    else if (x 2]) return ch(a, left, (left + right) / 2-1, x,len);
}


int main()
{
    int a[] = { 1,3,5,7,9,11,13,15,17,19 };
    cout "sizeof(a)/sizeof(a[0])= " sizeof(a) / sizeof(a[0]) " sizeof(a)=" sizeof(a)  endl;
    cout "result=" 0,9,0,sizeof(a)/sizeof(a[0]))  endl;
    for (auto i : a)
        cout " "  i;
}