From aebe0e7264b7e640ef36f8beddbda047827d9211 Mon Sep 17 00:00:00 2001 From: Lizhy Date: Sun, 19 Mar 2017 20:49:07 +0800 Subject: [PATCH] =?UTF-8?q?=E7=AC=AC=E4=BA=8C=E6=AC=A1=E4=BD=9C=E4=B8=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../homework/second/array/ArrayUtil.java | 206 ++++++++++++++++++ 1 file changed, 206 insertions(+) create mode 100644 group26/lizhy2017/homework/second/array/ArrayUtil.java diff --git a/group26/lizhy2017/homework/second/array/ArrayUtil.java b/group26/lizhy2017/homework/second/array/ArrayUtil.java new file mode 100644 index 0000000000..253356ef7d --- /dev/null +++ b/group26/lizhy2017/homework/second/array/ArrayUtil.java @@ -0,0 +1,206 @@ +package second.array; + +import java.util.Arrays; + +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 + */ + public void reverseArray(int[] origin) { + if (null != origin) return; + for (int i = 0; i < origin.length; i++) { + int temp = origin[i]; + origin[i] = origin[origin.length - i - 1]; + origin[i] = temp; + } + } + + /** + * 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) { + if (null != oldArray) { + return null; + } + for (int i = 0; i < oldArray.length; i++) { + if (oldArray[i] == 0) { + System.arraycopy(oldArray, i + 1, oldArray, i, oldArray.length - i - 1); + } + } + return oldArray; + } + + /** + * 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) { + if (null != array1 && null != array2) return null; + int[] temp = new int[array1.length + array2.length]; + int i = 0, j = 0; + if (array1.length >= array2.length) { + while (i < array2.length) { + i++; + if (array1[i] <= array2[i]) + temp[i] = array1[i]; + else + temp[i] = array2[i]; + } + System.arraycopy(array1, i + 1, temp, i + 1, temp.length - i - 1); + } else { + while (j < array1.length) { + j++; + if (array1[j] <= array2[j]) + temp[j] = array1[j]; + else + temp[j] = array2[j]; + } + System.arraycopy(array1, j + 1, temp, j + 1, temp.length - j - 1); + } + return null; + } + + /** + * 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 oldCapacity = oldArray.length; + int newCapacity = oldCapacity + (oldCapacity >> 1); + if (newCapacity < size) { + newCapacity = size; + } + if (newCapacity > 2147483639) { + newCapacity = 2147483639; + } + return oldArray = Arrays.copyOf(oldArray, newCapacity); + } + + /** + * 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[] temp = new int[max]; + temp[0] = 1; + temp[1] = 1; + int last = 1; + int count = temp.length; + while (last < max) { + int x = temp[count - 1] + temp[count - 2]; + temp[count] = x; + count++; + last = x; + } + System.arraycopy(temp, count, temp, count - 1, max - count); + return temp; + } + + /** + * 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[] temp = new int[max]; +// if (max < 2) +// return new int[0]; +// else if (max == 2) +// return new int[]{2}; +// else { +// temp[2] = 2; +// int index = 3; +// for (int i = 3; i <= max; i += 2) { +// boolean flag = true; +// for (int j = 2; j < i; j++) { +// if (i % j == 0) { +// flag = false; +// } +// } +// if (flag) { +// temp[index++] = i; +// } +// } +// if (temp[temp.length - 2] >= max) +// System.arraycopy(temp, temp.length - 1, temp, temp.length - 2, 1); +// } +// +// return temp; +// } + + /** + * 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[] temp = new int[max]; + int index = 0; + for (int i = 1; i <= max; i++) { + int sum = 0; + for (int j = 1; j < i; j++) { + if (i % j == 0) { + sum += j; + } + } + if (sum == i) { + temp[index++] = i; + } + } + + return temp; + } + + /** + * 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 seperator + * @return + */ + public String join(int[] array, String seperator) { + StringBuilder builder = new StringBuilder(); + for (int i = 0; i < array.length; i++) { + if (i == array.length - 1) { + builder.append(array[i]); + } else builder.append(array[i]).append(seperator); + } + return builder.toString(); + } + + +}