交作业0319

group24/1148285693/learning2017/learning-basic/src/main/java/me/lzb/homework0312/basic/ArrayList.java

@@ -12,13 +12,14 @@ public class ArrayList implements List {
 
 
     public void add(Object o) {
-        if (elementData.length < size + 1) {
-            Object[] target = new Object[size + 1];
-            System.arraycopy(elementData, 0, target, 0, elementData.length);
-            elementData = target;
-        }
-        elementData[size] = o;
-        size = size + 1;
+//        if (elementData.length < size + 1) {
+//            Object[] target = new Object[size + 1];
+//            System.arraycopy(elementData, 0, target, 0, elementData.length);
+//            elementData = target;
+//        }
+//        elementData[size] = o;
+//        size = size + 1;
+        add(size, o);
     }
 
 
@@ -29,7 +30,7 @@ public void add(int index, Object o) throws IndexOutOfBoundsException {
 
         int leftSize = index;
         int rightSize = size - index;
-        Object[] target = new Object[elementData.length + 1];
+        Object[] target = new Object[size + 1];
         System.arraycopy(elementData, 0, target, 0, leftSize);
         target[index] = o;
         System.arraycopy(elementData, index, target, index + 1, rightSize);

group24/1148285693/learning2017/learning-basic/src/main/java/me/lzb/homework0319/array/ArrayUtil.java

@@ -0,0 +1,337 @@
+package me.lzb.homework0319.array;
+
+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[] target = new int[origin.length];
+        for (int i = 0; i < origin.length; i++) {
+            target[i] = origin[origin.length - 1 - i];
+        }
+        origin = target;
+    }
+
+    /**
+     * 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 l = 0;
+        for (int i = 0; i < oldArray.length; i++) {
+            if (oldArray[i] != 0) {
+                l = l + 1;
+            }
+        }
+
+        int[] target = new int[l];
+
+        int a = 0;
+        for (int i = 0; i < oldArray.length; i++) {
+            if (oldArray[i] != 0) {
+                target[a] = oldArray[i];
+                a = a + 1;
+            }
+        }
+
+        return target;
+    }
+
+    /**
+     * 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
+     */
+
+    //一个一个放进去，循环次数有点多
+    public int[] merge(int[] array1, int[] array2) {
+
+
+        int[] tmp = new int[array1.length + array2.length];
+
+
+        int mini = 0;
+        int a1 = array1[0];
+        int a2 = array2[0];
+
+        if(a1 < a2){
+            mini = a1;
+        }else {
+            mini = a2;
+        }
+
+        tmp[0] = mini;
+
+        int l3 = 0;
+
+        for (int i = 1; i < array1.length + array2.length; i++) {
+
+//            if(mini >= array1[l1 - 1] && mini >= array2[l2 - 1]){
+//                l3 = i;
+//                break;
+//            }
+
+            int oldMin = mini;
+
+
+
+            int aa1 = mini;
+            if(mini < array1[array1.length - 1] ){
+                for (int j = 0; j < array1.length; j++) {
+                    if(array1[j] > mini){
+                        aa1 = array1[j];
+                        break;
+                    }
+                }
+
+            }
+
+            int aa2 = mini;
+            if(mini < array2[array2.length - 1] ){
+                for (int j = 0; j < array2.length; j++) {
+                    if(array2[j] > mini){
+                        aa2 = array2[j];
+                        break;
+                    }
+                }
+            }
+
+
+            if(aa1 != oldMin && aa2 != oldMin){
+                if(aa1 < aa2){
+                    mini = aa1;
+                }else {
+                    mini = aa2;
+                }
+            }else if(aa1 != oldMin){
+                mini = aa1;
+            }else {
+                mini = aa2;
+            }
+
+
+            if(oldMin == mini){
+                l3 = i;
+                break;
+            }
+
+            tmp[i] = mini;
+        }
+
+        int[] result = new int[l3];
+
+        System.arraycopy(tmp, 0, result, 0, l3);
+
+
+
+        return result;
+    }
+
+
+
+
+    /**
+     * 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[] = new int[oldArray.length + size];
+
+        System.arraycopy(oldArray, 0, newArray, 0, oldArray.length);
+        return 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) {
+        if (max <= 1){
+            return new int[0];
+        }
+
+        int[] result = {1, 1};
+
+
+        int i = 2;
+
+        int n = 0;
+
+        while (n < max){
+            int[] t = new int[result.length + 1];
+            System.arraycopy(result, 0, t, 0, result.length);
+            n = t[i-1] + t[i - 2];
+
+            if(n >= max){
+                return result;
+            }
+
+            t[i] = n;
+
+            result = t;
+            i = i + 1;
+        }
+
+        return null;
+    }
+
+
+    /**
+     * 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) {
+        if (max <= 2){
+            return new int[0];
+        }
+
+        if (max == 3){
+            return new int[]{2};
+        }
+
+
+        int[] primes = new int[max+1];
+        primes[0] = 2;
+        int count = 1;
+        for (int i = 3; i < max; i = i + 2) {
+
+            boolean isPrime = true;
+            for (int j = 3; j < i; j++) {
+                if(i % j == 0){
+                    isPrime = false;
+                    break;
+                }
+            }
+
+            if(isPrime){
+                primes[count] = i;
+                count = count + 1;
+            }
+        }
+
+        int[] result = new int[count];
+        System.arraycopy(primes, 0, result, 0, count);
+
+        return result;
+
+    }
+
+    private boolean isPrime(int a){
+        if (a < 2) {
+            return false;
+        }
+
+        if (a == 2) {
+            return true;
+        }
+
+        if(a % 2 == 0){
+            return false;
+        }
+
+
+        for (int i = 3; i < a; i = i + 2) {
+            if(a % i == 0){
+                return false;
+            }
+        }
+
+        return true;
+    }
+
+
+
+    /**
+     * 所谓“完数”， 是指这个数恰好等于它的真因子之和，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) {
+        if (max < 6){
+            return new int[0];
+        }
+
+
+        int[] pns = new int[max];
+
+        int count = 0;
+        for (int i = 6; i < max; i++) {
+            if (isPerfectNumber(i)){
+                pns[count] = i;
+                count = count + 1;
+            }
+        }
+
+
+
+        int[] result = new int[count];
+        System.arraycopy(pns, 0, result, 0, count);
+        return result;
+    }
+
+
+    private boolean isPerfectNumber(int a){
+        if(a < 6){
+            return false;
+        }
+
+        int sum = 0;
+        for (int i = 1; i < a; i++) {
+            if(a % i == 0){
+                sum = sum + i;
+            }
+        }
+
+        return sum == a;
+    }
+
+
+
+    /**
+     * 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
+     * @return
+     */
+    public static  String join(int[] array, String seperator) {
+        String result = "";
+        for (int i = 0; i < array.length; i++) {
+            result = result + array[i] + seperator ;
+        }
+
+        result = result.substring(0, result.length() - 1);
+        return result;
+    }
+
+}