/** * 问题点: 没写注释,代码比较难读。尤其 merge方法。 */ package com.github.eulerlcs.jmr.algorithm; 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 * @return */ public static void reverseArray(int[] origin) { if (origin == null || origin.length < 2) { return; } for (int head = 0, tail = origin.length - 1; head < tail; head++, tail--) { origin[head] = origin[head] ^ origin[tail]; origin[tail] = origin[head] ^ origin[tail]; origin[head] = origin[head] ^ origin[tail]; } } /** * 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 static int[] removeZero(int[] oldArray) { if (oldArray == null) { return new int[0]; } int count = 0; for (int i = 0; i < oldArray.length; i++) { if (oldArray[i] != 0) count++; } int[] newArray = new int[count]; int newIndex = 0; for (int i = 0; i < oldArray.length; i++) { if (oldArray[i] != 0) { newArray[newIndex] = oldArray[i]; newIndex++; } } return 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 static int[] merge(int[] array1, int[] array2) { if (array1 == null || array1.length == 0) { if (array2 == null || array2.length == 0) { return new int[0]; } else { return Arrays.copyOf(array2, array2.length); } } else if (array2 == null || array2.length == 0) { return Arrays.copyOf(array1, array1.length); } int[] result = new int[array1.length + array2.length]; int idxResult = 0; int idx1 = 0; int idx2 = 0; for (; idx1 < array1.length; idx1++) { if (array1[idx1] < array2[idx2]) { result[idxResult] = array1[idx1]; idxResult++; } else if (array1[idx1] == array2[idx2]) { result[idxResult] = array1[idx1]; idxResult++; idx2++; } else { for (; idx2 < array2.length; idx2++) { if (array2[idx2] < array1[idx1]) { result[idxResult] = array2[idx2]; idxResult++; } else { if (array2[idx2] == array1[idx1]) { idx2++; } break; } } if (idx2 == array2.length) { break; } else { idx1--; } } } if (idx1 < array1.length) { System.arraycopy(array1, idx1, result, idxResult, array1.length - idx1); idxResult += array1.length - idx1; } if (idx2 < array2.length) { System.arraycopy(array2, idx2, result, idxResult, array2.length - idx2); idxResult += array2.length - idx2; } result = removeZero(result); 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 increaseCapacity * @return */ public static int[] grow(int[] oldArray, int increaseCapacity) { if (oldArray == null || increaseCapacity < 0) { return new int[0]; } int newCapacity = oldArray.length + increaseCapacity; int[] newArray = Arrays.copyOf(oldArray, newCapacity); 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 static int[] fibonacci(int max) { if (max <= 1) { return new int[0]; } int[] result = new int[10]; result[0] = 1; result[1] = 1; int idx = 2; int sum = 2; while (sum < max) { if (idx >= result.length) { grow(result, result.length * 2); } result[idx] = sum; sum = result[idx - 1] + result[idx]; idx++; } result = removeZero(result); return result; } /** * 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 static int[] getPrimes(int max) { if (max < 2) { return new int[0]; } int[] all = new int[max]; int index = 0; int temp = 0; for (int i = 0; i < max; i++) { all[i] = i; } all[0] = 0; all[1] = 0; index = 2; // 筛法 loops: for (; index < max;) { for (int i = 2;; i++) { temp = index * i; if (temp >= max) { break; } all[temp] = 0; } for (int i = index + 1; i < max; i++) { if (all[i] != 0) { index = i; continue loops; } } break; } int[] result = removeZero(all); return result; } /** * 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 static long[] getPerfectNumbers(long max) { long[] perfect = new long[49];// 到2016年1月为止,共发现了49个完全数 int idx = 0; long sum = 1; long sqrt = 0; for (long n = 2; n < max; n++) { sum = 1; sqrt = (long) Math.sqrt(n); for (long i = 2; i <= sqrt; i++) { if (n % i == 0) sum += i + n / i; } if (sum == n) { perfect[idx] = n; idx++; } } // return removeZero(perfect); return perfect; } /** * 用separator 把数组 array给连接起来 例如array= [3,8,9], separator = "-" 则返回值为"3-8-9" * * @param array * @param separator * @return */ public static String join(int[] array, String separator) { if (array == null || array.length == 0) { return ""; } StringBuilder sb = new StringBuilder(); for (int i = 0; i < array.length - 1; i++) { sb.append(array[i] + separator); } sb.append(String.valueOf(array[array.length - 1])); return sb.toString(); } }