二维数组面试题（外加创建动态数组）

自己老是忘掉，所以干脆写一个在博客存着。另外，附上两道题：1 二维数组中的查值 2 顺时针打印一个二维数组C++创建动态数组：一维数组：int *x = new int [size];delete [] x;二维数组，直接上代码（包括如何把二维数组作为参数传递）：#include <iostream>using namespace std;int colFind(int **a, int row, int col){for (int i = 0; i != row; ++i)for (int j = 0; j != col; ++j)cout << a[i][j] << endl;return 0;}

int main()
{
int **x = new int*[5];//分配空间
for (int i = 0; i != 5; ++i)
x[i] = new int[5];

for (int i = 0; i != 5; ++i) //数组赋值
for (int j = 0; j != 5; ++j)
x[i][j] = i * 5 + j;

colFind(x, 5, 5);//打印数组

for (int i = 0; i != 5; ++i)//撤销空间
delete[] x[i];
delete[] x;

return 0;
}

两道算法题

第一道：返回一个key元素下标的pair对，如果没找到则返回（-1，-1）。

#include <iostream>
using namespace std;

pair<int, int> FindInMat(int **x , int row, int col, int key)
{
int i = 0, j = col - 1;
while (i < row && j >= 0)
{
if (x[i][j] == key)
return make_pair(i, j);
else if (x[i][j] > key)
--j;
else
++i;
}
return make_pair(-1, -1);
}

//测试代码：int main(){int **x = new int* [4];x[0] = new int[4]{1, 2, 8, 9};x[1] = new int[4]{2, 4, 9, 12};x[2] = new int[4]{4, 7, 10, 13};x[3] = new int[4]{6, 8, 11, 15};//colFind(x, 4, 4);cout << FindInMat(x, 4, 4, 11).first << FindInMat(x, 4, 4, 11).second;system("pause");return 0;}

第二道：环形打印一个二维数组

<pre class="cpp" name="code">#include <iostream>using namespace std;//打印一个圈void Print(int **a, int row, int col, int start){int stopX = row - start - 1;int stopY = col - start - 1;if ((stopX == start) && (stopY == start))cout << a[start][start] << endl;else{for (int i = start; i <= stopX; ++i)cout << a[start][i] << endl;for (int i = start + 1; i <= stopY; ++i)cout << a[i][stopX] << endl;for (int i = stopX - 1; i >= start; --i)cout << a[stopY][i] << endl;if (stopY  - start >= 2)for (int i = stopY -1; i >= start + 1; --i)cout << a[i][start] << endl;}}//打印函数的实体void PrintCircle(int **a, int row, int col){if (row <= 0 || col <= 0 || a == NULL)return;else{int start = 0;while (row > 2 * start && col > 2 * start){Print(a, row, col, start);++start;}}}//测试代码int main(){int **a = new int*[5];for (int i = 0; i != 5; ++i){a[i] = new int[5];}for (int i = 0; i != 5; ++i)for (int j = 0; j != 5; ++j)a[i][j] = i * 5 + j;PrintCircle(a, 5, 5);system("pause");return 0;}