Merge pull request #42 from McRays/master

史磊2017.3.5提交作业
onlyliuxin · Mar 5, 2017 · 3376af0 · 3376af0
2 parents fcad223 + 34686e9
commit 3376af0
Show file tree

Hide file tree

Showing 7 changed files with 732 additions and 0 deletions.
diff --git a/group08/782476895/20170305/ArrayUtil.java b/group08/782476895/20170305/ArrayUtil.java
@@ -0,0 +1,314 @@
+package com.sl.test20170304.arrayutil;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+import org.junit.Test;
+
+public class ArrayUtil {
+
+	/**
+	 * 给定一个整形数组a , 对该数组的值进行置换
+		例如： a = [7, 9 , 30, 3]  ,   置换后为 [3, 30, 9,7]
+		如果     a = [7, 9, 30, 3, 4] , 置换后为 [4,3, 30 , 9,7]
+	 * @param origin
+	 * @return
+	 */
+	public void reverseArray(int[] origin){
+		int low = 0;
+		int high = origin.length - 1;
+		while(low < high){
+			int temp = 0;
+			temp = origin[low];
+			origin[low] = origin[high];
+			origin[high] = temp;
+			low++;
+			high--;
+		}
+	}
+
+	@Test
+	public void test1(){
+		int[] arr = {7, 9 , 30, 3};
+		reverseArray(arr);
+		System.out.println(Arrays.toString(arr));
+	}
+
+	/**
+	 * 现在有如下的一个数组：   int oldArr[]={1,3,4,5,0,0,6,6,0,5,4,7,6,7,0,5}   
+	 * 要求将以上数组中值为0的项去掉，将不为0的值存入一个新的数组，生成的新数组为：   
+	 * {1,3,4,5,6,6,5,4,7,6,7,5}  
+	 * @param oldArray
+	 * @return
+	 */
+
+	public int[] removeZero(int[] oldArray){
+		int[] arr = new int[oldArray.length];
+		int num = 0;
+		for(int i = 0;i < oldArray.length;i++){
+			if(oldArray[i] != 0){
+				arr[num] = oldArray[i];
+				num++;
+			}
+		}
+		int[] newArray = Arrays.copyOf(arr, num);
+		return newArray;
+	}
+
+	@Test
+	public void test2(){
+		int oldArr[] = {1,3,4,5,0,0,6,6,0,5,4,7,6,7,0,5};
+		int[] newArray = removeZero(oldArr);
+		System.out.println(Arrays.toString(newArray));
+	}
+
+
+
+	/**
+	 * 给定两个已经排序好的整形数组， a1和a2 ,  创建一个新的数组a3, 使得a3 包含a1和a2 的所有元素， 并且仍然是有序的
+	 * 例如 a1 = [3, 5, 7,8]   a2 = [4, 5, 6,7]    则 a3 为[3,4,5,6,7,8]    , 注意： 已经消除了重复
+	 * @param array1
+	 * @param array2
+	 * @return
+	 */
+
+	public int[] merge(int[] array1, int[] array2){
+		int len1 = array1.length;
+		int len2 = array2.length;
+
+		int[] arr = new int[len1 + len2]; 
+		int len = 0;
+		int i = 0;
+		int j = 0;
+		int min = 0;
+		while(i < len1 && j< len2){
+			if(array1[i] == array2[j]){
+				min = array1[i];
+				arr[len] = min;
+				i++;
+				j++;
+				len++;
+				continue;
+			}
+
+			if(array1[i] > array2[j]){
+				min = array2[j];
+				arr[len] = min;
+				j++;
+				len++;
+				continue;
+			}
+
+			if(array1[i] < array2[j]){
+				min = array1[i];
+				arr[len] = min;
+				i++;
+				len++;
+				continue;
+			}
+		}
+
+		if(i > j){
+			for(int k = j;k < len2;k++){
+				arr[len] = array2[k];
+				len++;
+			}
+		}
+
+		if(i < j){
+			for(int k = i;k < len1;k++){
+				arr[len] = array1[k];
+				len++;
+			}
+		}
+
+		int[] newArray = Arrays.copyOf(arr, len);
+		return  newArray;
+	}
+
+
+	@Test
+	public void test3(){
+		int[] a1 = {3, 7, 9,10,34,78,123};
+		int[] a2 = {4, 5, 6,7,56,34,345,2432,545345345};
+		int[] newArray = merge(a1, a2);
+		System.out.println(Arrays.toString(newArray));
+	}
+
+
+
+	/**
+	 * 把一个已经存满数据的数组 oldArray的容量进行扩展， 扩展后的新数据大小为oldArray.length + size
+	 * 注意，老数组的元素在新数组中需要保持
+	 * 例如 oldArray = [2,3,6] , size = 3,则返回的新数组为
+	 * [2,3,6,0,0,0]
+	 * @param oldArray
+	 * @param size
+	 * @return
+	 */
+	public int[] grow(int [] oldArray,  int size){
+		int[] newArray = Arrays.copyOf(oldArray, oldArray.length + size);
+		return newArray;
+	}
+
+
+	@Test
+	public void test4(){
+		int[] oldArray = {2,3,6};
+		int[] newArray = grow(oldArray,3);
+		System.out.println(Arrays.toString(newArray));
+ 	}
+
+	/**
+	 * 斐波那契数列为：1，1，2，3，5，8，13，21......  ，给定一个最大值， 返回小于该值的数列
+	 * 例如， max = 15 , 则返回的数组应该为 [1，1，2，3，5，8，13]
+	 * max = 1, 则返回空数组 []
+	 * @param max
+	 * @return
+	 */
+	public int[] fibonacci(int max){
+
+		int[] arr = new int[100];
+		int len = 2;
+		int f1 = 1;
+		int f2 = 1;
+		int f3 = f1 + f2;
+
+		if(max == 1){
+			return new int[0];
+		}
+
+		arr[0] = f1;
+		arr[1] = f2;
+
+		while(f3 <= max){
+			arr[len] = f3;
+			len++;
+			f1 = f2;
+			f2 = f3;
+			f3 = f1 + f2;
+		}
+		int[] newArray = Arrays.copyOf(arr, len);
+		return newArray;
+	}
+
+	@Test
+	public void test5(){
+		int[] newArray = fibonacci(100);
+		System.out.println(Arrays.toString(newArray));
+	}
+
+
+	/**
+	 * 返回小于给定最大值max的所有素数数组
+	 * 例如max = 23, 返回的数组为[2,3,5,7,11,13,17,19]
+	 * @param max
+	 * @return
+	 */
+	public int[] getPrimes(int max){
+		int[] arr = new int[max];
+		int len = 0; 
+		for(int i = 0;i <= max;i++){
+			if(i == 0 || i == 1){
+				continue;
+			}
+			if(isPrime(i)){
+				arr[len] = i;
+				len++;
+			}
+		}
+		int[] newArray = Arrays.copyOf(arr, len);
+		return newArray;
+	}
+
+	private boolean isPrime(int n){
+	    for(int i=2;i<=n/2;i++){
+	         if(n%i == 0)
+	              return false; 
+	    }
+	    return true; 
+	}
+
+	@Test
+	public void test6(){
+		int max = 100;
+		int[] newArray = getPrimes(max);
+		System.out.println(Arrays.toString(newArray));
+	}
+
+	/**
+	 * 所谓“完数”， 是指这个数恰好等于它的因子之和，例如6=1+2+3
+	 * 给定一个最大值max， 返回一个数组， 数组中是小于max 的所有完数
+	 * @param max
+	 * @return
+	 */
+	public int[] getPerfectNumbers(int max){
+		int[] arr = new int[max];
+		int len = 0; 
+		for(int i = 0;i <= max;i++){
+			if(i == 0 || i == 1){
+				continue;
+			}
+			if(isPerfectNumber(i)){
+				arr[len] = i;
+				len++;
+			}
+		}
+		int[] newArray = Arrays.copyOf(arr, len);
+		return newArray;
+	}
+
+
+	private boolean isPerfectNumber(int num){
+		List<Integer> list = new ArrayList<Integer>();
+		int sum = 0;
+		for(int i = 1;i < num;i++){
+			if(num % i == 0){
+				list.add(i);
+			}
+		}
+
+		for(int j : list){
+			sum += j;
+		}
+
+		return sum == num ? true : false;
+	}
+
+
+	@Test
+	public void test7(){
+		int max = 100000;
+		int[] newArray = getPerfectNumbers(max);
+		System.out.println(Arrays.toString(newArray));
+	}
+
+
+	/**
+	 * 用seperator 把数组 array给连接起来
+	 * 例如array= [3,8,9], seperator = "-"
+	 * 则返回值为"3-8-9"
+	 * @param array
+	 * @param s
+	 * @return
+	 */
+	public String join(int[] array, String seperator){
+		StringBuilder sb = new StringBuilder();
+		for(int num : array){
+			sb.append(num).append(seperator);
+		}
+
+		String str = sb.toString();
+		str = str.substring(0, str.lastIndexOf(seperator));
+		return str;
+	}
+
+	@Test
+	public void test8(){
+		int[] arr = {1,2,5,6,78,34,332,4452,4342,3,435,233};
+		System.out.println(join(arr,"-"));
+	}
+
+
+}
diff --git a/group08/782476895/20170305/struts/LoginAction.java b/group08/782476895/20170305/struts/LoginAction.java
@@ -0,0 +1,41 @@
+package com.sl.test20170304.struts;
+
+/**
+ * 这是一个用来展示登录的业务类， 其中的用户名和密码都是硬编码的。
+ * @author liuxin
+ *
+ */
+public class LoginAction{
+    private String name ;
+    private String password;
+    private String message;
+
+    public String getName() {
+        return name;
+    }
+
+    public String getPassword() {
+        return password;
+    }
+
+
+
+	public String execute(){
+            if("test".equals(name) && "1234".equals(password)){
+                this.message = "login successful";
+                return "success";
+            }
+            this.message = "login failed,please check your user/pwd";
+            return "fail";
+    }
+
+    public void setName(String name){
+        this.name = name;
+    }
+    public void setPassword(String password){
+        this.password = password;
+    }
+    public String getMessage(){
+        return this.message;
+    }
+}