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honokaBiuhonokaBiu
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Merge pull request onlyliuxin#28 from 592146505/master
实现简单的struts和数组操作
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‎group20/592146505/array_util/src/org/wsc/array/ArrayUtil.java

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‎group20/592146505/array_util/src/org/wsc/array/ArrayUtilTest.java

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‎group20/592146505/coderising/src/org/wsc/array/ArrayUtil.java

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package org.wsc.array;
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public class ArrayUtil {
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/**
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* 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 =
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* [7, 9, 30, 3, 4] , 置换后为 [4,3, 30 , 9,7]
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*
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* @param origin
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* @return
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*/
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public static void reverseArray(int[] origin) {
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// 折半
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for (int i = 0; i < (origin.length >> 1); i++) {
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int num = origin[i];
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origin[i] = origin[origin.length - 1 - i];
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origin[origin.length - 1 - i] = num;
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}
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}
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/**
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* 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}
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* 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}
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*
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* @param oldArray
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* @return
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*/
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public static int[] removeZero(int[] oldArray) {
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int count = 0;// 计数器
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/*
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* 利用冒泡，将0元素向后排 {1,3,4,5,0,0,6,6,0,5,4,7,6,7,0,5}
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* {1,3,4,5,0,6,6,0,5,4,7,6,7,0,5,0} {1,3,4,5,6,6,0,5,4,7,6,7,0,5,0,0}
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* ....
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*/
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for (int i = 0; i < oldArray.length - count; i++) {
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// 索引为i的元素为0，则依次将索引i的元素与i+1的元素对换
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if (oldArray[i] == 0) {
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for (int j = i; j < oldArray.length - 1 - count; j++) {
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int num = oldArray[j];
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oldArray[j] = oldArray[j + 1];
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oldArray[j + 1] = num;
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}
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count++;// 计数器+1
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i--;// 防止原索引i+1位置的元素为0,
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}
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}
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// 创建新数组
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int[] newArray = new int[oldArray.length - count];
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System.arraycopy(oldArray, 0, newArray, 0, newArray.length);
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return newArray;
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}
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/**
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* 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}
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* 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}
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*
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* @param oldArray
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* @return
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*/
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public static int[] removeZero2(int[] oldArray) {
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int count = 0;// 计数器
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for (int i = 0; i < oldArray.length; i++) {
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if (oldArray[i] == 0)
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count++;// 计数器+1
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}
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// 创建新数组
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int[] newArray = new int[oldArray.length - count];
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for (int i = 0, j = 0; i < oldArray.length; i++) {
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if (oldArray[i] != 0) {
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newArray[j] = oldArray[i];
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j++;
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}
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}
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return newArray;
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}
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/**
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* 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 =
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* [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
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*
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* @param array1
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* @param array2
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* @return
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*/
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public static int[] merge(int[] array1, int[] array2) {
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int[] newArray = new int[array1.length + array2.length];
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int i = 0, j = 0, k = 0;
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while (i < array1.length && j < array2.length) {
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// <= 都取 array1
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if (array1[i] <= array2[j]) {
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// 等于时，将array2下标++
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if (array1[i] == array2[j])
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j++;
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newArray[k++] = array1[i++];
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} else
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newArray[k++] = array2[j++];
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}
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// 将没有循环完毕的元素插入
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while (i < array1.length)
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newArray[k++] = array1[i++];
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while (j < array2.length)
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newArray[k++] = array2[j++];
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int[] result = newArray;
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// 长度缩短则新建数组
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if (k < newArray.length) {
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result = new int[k];
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for (int l = 0; l < result.length; l++)
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result[l] = newArray[l];
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}
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return result;
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}
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/**
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* 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
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* 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
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* [2,3,6,0,0,0]
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*
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* @param oldArray
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* @param size
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* @return
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*/
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public static int[] grow(int[] oldArray, int size) {
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int[] newArray = new int[oldArray.length + size];
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for (int i = 0; i < oldArray.length; i++) {
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newArray[i] = oldArray[i];
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}
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return newArray;
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}
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/**
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* 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 ,
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* 则返回的数组应该为 [1，1，2，3，5，8，13] max = 1, 则返回空数组 []
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*
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* @param max
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* @return
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*/
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public static int[] fibonacci(int max) {
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if (max <= 1)
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return new int[] {};
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int[] nums = new int[max];
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nums[0] = nums[1] = 1;
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int flag;
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for (flag = 0; (flag < max - 2 && nums[flag] + nums[flag + 1] < max); flag++) {
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nums[flag + 2] = nums[flag] + nums[flag + 1];
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}
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// 创建新数组
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int[] newArray = nums;
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if (newArray.length != flag + 2) {
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newArray = new int[flag + 2];
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for (int i = 0; i < newArray.length; i++) {
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newArray[i] = nums[i];
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}
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}
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return newArray;
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}
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/**
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* 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]
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*
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* @param max
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* @return
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*/
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public static int[] getPrimes(int max) {
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int[] array = new int[max>>1];
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int flag = 0;
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for (int i = 2; i < max; i++) {
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int j;
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for (j = 2; j <= (i >> 1); j++) {
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if (i % j == 0)
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break;
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}
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//如果大于，则证明j++有运行，已经完整对比
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if(j > i>>1)
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array[flag++] = i;
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}
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int[] newArray = array;
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if(flag < array.length){
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newArray = new int[flag];
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for (int i = 0; i < newArray.length; i++) {
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newArray[i] = array[i];
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}
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}
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return newArray;
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}
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/**
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* 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
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*
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* @param max
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* @return
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*/
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public static int[] getPerfectNumbers(int max) {
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int[] array = new int[max];
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int flag = 0;
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for (int i = 1; i < max; i++) {
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int sum = 0;
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for (int j = 1; j < i; j++) {
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if (i % j == 0)
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sum+=j;
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}
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//如果大于，则证明j++有运行，已经完整对比
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if(sum == i)
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array[flag++] = i;
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}
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int[] newArray = array;
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if(flag < array.length){
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newArray = new int[flag];
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for (int i = 0; i < newArray.length; i++) {
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newArray[i] = array[i];
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}
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}
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return newArray;
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}
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/**
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* 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"
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*
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* @param array
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* @param s
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* @return
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*/
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public static String join(int[] array, String seperator) {
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String str = "";
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for (int i = 0; i < array.length; i++)
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str += i != array.length - 1 ? array[i] + seperator : array[i];
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return str;
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}
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}

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