第二次任务

2020-Autumn/Backend/task_02/Zeppel-Tao/README.md

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+# 第二次任务总结
+
+**1、学习和使用指针与数组**
+
+**2、学习了时间复杂度与空间复杂度的概念**
+
+**3、学习了动态数组的概念**
+
+**4、了解了递归的概念**
+
+*****
+
+## 出现的问题
+
+**1、开始时未弄明白动态数组的概念**
+
+**2、多重指针的含义**
+
+**3、不知该怎样去优化第一题的复杂度问题**
+
+
+
+
+

2020-Autumn/Backend/task_02/Zeppel-Tao/第二次任务.cpp

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+#include <iostream>
+using namespace std;
+//第一道
+void text01(int m, int n, int q, int* arrA, int* arrB)
+{
+	int i, j;
+	i = m - 1;
+	j = n - 1;
+	while (i >= 0 && j >= 0)
+	{
+		if (arrA[i] >= arrB[j]) //从后向前排序，将较大的值赋给A。
+		{
+			arrA[i + j + 1] = arrA[i];
+			i--;
+		}
+		else
+		{
+			arrA[i + j + 1] = arrB[j];
+			j--;
+		}
+		while (j >= 0)//当B中有剩余的值时，将剩余的值赋到A中
+			          //若剩余的是A中的值，则直接在数组A中
+		{
+			arrA[j] = arrB[j];
+		}
+
+	}
+}
+//第二道
+void text02(int** matrix, int n)
+{
+    matrix = new int* [n];
+    for (int i = 0; i < n; i++)
+    {
+        matrix[i] = new int[n];
+    }
+    cout << "翻转前:" << endl;
+    for (int i = 0; i < n; i++)
+
+    {
+        for (int j = 0; j < n; j++)
+        {
+
+            matrix[i][j] = i * n + j + 1;
+            cout << matrix[i][j] << "\t";
+        }
+        cout << endl;
+    }
+    for (int i = 0; i < n / 2; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            int temp = matrix[i][j];
+            matrix[i][j] = matrix[n - i - 1][j];
+            matrix[n - i - 1][j] = temp;
+
+        }
+    }
+    //对角线翻转
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < i; j++)
+        {
+            int temp = matrix[i][j];
+            matrix[i][j] = matrix[j][i];
+            matrix[j][i] = temp;
+        }
+    }
+    cout << "翻转后:" << endl;
+    for (int i = 0; i < n; i++)
+
+    {
+        for (int j = 0; j < n; j++)
+        {
+
+            cout << matrix[i][j] << "\t";
+        }
+        cout << endl;
+    }
+}
+//第三道
+void text03(int* a, int n)
+{
+    int size = 10000;
+    int* arr = new int[size];
+    int n = 0;
+    cin >> n;
+    for (int i = 0; i < n; i++)
+    {
+        arr[i] = i;
+        cout << arr[i] << " ";
+    }
+}
+//第四道
+int* text04(int n)
+{
+    cin >> n;
+    int* arr = new int[n];
+    for (int i = 0; i < n; i++)
+    {
+        arr[i] = i + 1;
+    }
+
+    for (int i = 0; i < n; i++)
+    {
+        cout << arr[i] << "\t";
+    }
+    return arr;
+}
+//第五道
+int** text05(int m,int n)
+{
+    int m = 0;
+    int n = 0;
+    cout << "请输入排数: ";
+    cin >> m;
+    cout << "请输入列数: ";
+    cin >> n;
+    int** arr = new int* [m];
+    for (int i = 0; i < m; i++)
+    {
+        arr[i] = new int[n];
+    }
+    for (int i = 0; i < m; i++)
+
+    {
+        for (int j = 0; j < n; j++)
+        {
+            arr[i][j] = i * n + j;  //若使用指针则为*(*(arr+i)+j) cin >> arr[i][j]  
+            cout << arr[i][j] << "\t";
+        }
+        cout << endl;
+    }
+    for (int i = 0; i < m; i++)
+    {
+        delete[] arr[i];
+    }
+    delete[] arr;
+}
+//第六道
+int climbstairs(int n)
+{
+    if (n == 0)
+    {
+        return 0;
+    }
+    else if (n == 1)
+    {
+        return 1;
+    }
+    else if (n == 2)
+    {
+        return 2;
+    }
+    else {
+        return climbstairs(n - 1) + climbstairs(n - 2); //使用递归
+    }
+}