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| 1 | +package array; |
| 2 | + |
| 3 | +public class ArrayUtil { |
| 4 | + |
| 5 | + /** |
| 6 | + * 给定一个整形数组a , 对该数组的值进行置换 |
| 7 | + 例如: a = [7, 9 , 30, 3] , 置换后为 [3, 30, 9,7] |
| 8 | + 如果 a = [7, 9, 30, 3, 4] , 置换后为 [4,3, 30 , 9,7] |
| 9 | + * @param origin |
| 10 | + * @return |
| 11 | + */ |
| 12 | + public void reverseArray(int[] origin){ |
| 13 | + if (origin.length == 0) { |
| 14 | + return; |
| 15 | + } |
| 16 | + int n = origin.length - 1; |
| 17 | + int temp; |
| 18 | + for (int i = 0; i < n/2; i++) { |
| 19 | + temp = origin[i]; |
| 20 | + origin[i] = origin[n - i]; |
| 21 | + origin[n - i] = temp; |
| 22 | + } |
| 23 | + } |
| 24 | + |
| 25 | + /** |
| 26 | + * 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} |
| 27 | + * 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: |
| 28 | + * {1,3,4,5,6,6,5,4,7,6,7,5} |
| 29 | + * @param oldArray |
| 30 | + * @return |
| 31 | + */ |
| 32 | + |
| 33 | + public int[] removeZero(int[] oldArray){ |
| 34 | + if (oldArray.length == 0) { |
| 35 | + return null; |
| 36 | + } |
| 37 | + int n = oldArray.length; |
| 38 | + int zeros = 0; |
| 39 | + for (int i = 0; i < n - 1; i++) { |
| 40 | + if (oldArray[i] == 0) { |
| 41 | + zeros++; |
| 42 | + } |
| 43 | + } |
| 44 | + int[] result = new int[n - zeros]; |
| 45 | + int j = 0; |
| 46 | + for (int i = 0; i < n - zeros - 1; i++) { |
| 47 | + while (oldArray[j] == 0) { |
| 48 | + j++; |
| 49 | + } |
| 50 | + result[i] = oldArray[j]; |
| 51 | + j++; |
| 52 | + } |
| 53 | + return result; |
| 54 | + } |
| 55 | + |
| 56 | + /** |
| 57 | + * 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 |
| 58 | + * 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 |
| 59 | + * @param array1 |
| 60 | + * @param array2 |
| 61 | + * @return |
| 62 | + */ |
| 63 | + |
| 64 | + public int[] merge(int[] array1, int[] array2){ |
| 65 | + return null; |
| 66 | + } |
| 67 | + /** |
| 68 | + * 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 |
| 69 | + * 注意,老数组的元素在新数组中需要保持 |
| 70 | + * 例如 oldArray = [2,3,6] , size = 3,则返回的新数组为 |
| 71 | + * [2,3,6,0,0,0] |
| 72 | + * @param oldArray |
| 73 | + * @param size |
| 74 | + * @return |
| 75 | + */ |
| 76 | + public int[] grow(int [] oldArray, int size){ |
| 77 | + return null; |
| 78 | + } |
| 79 | + |
| 80 | + /** |
| 81 | + * 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 |
| 82 | + * 例如, max = 15 , 则返回的数组应该为 [1,1,2,3,5,8,13] |
| 83 | + * max = 1, 则返回空数组 [] |
| 84 | + * @param max |
| 85 | + * @return |
| 86 | + */ |
| 87 | + public static int[] fibonacci(int max){ |
| 88 | + if (max <= 1) { |
| 89 | + return new int[0]; |
| 90 | + } |
| 91 | + int n = 1; |
| 92 | + int init = helpFibonacci(n); |
| 93 | + while (init < max) { |
| 94 | + n++; |
| 95 | + init = helpFibonacci(n); |
| 96 | + } |
| 97 | + int[] result = new int[n - 1]; |
| 98 | + for (int i = 0; i < n - 1; i++) { |
| 99 | + result[i] = helpFibonacci(i + 1); |
| 100 | + } |
| 101 | + return result; |
| 102 | + } |
| 103 | + |
| 104 | + public static int helpFibonacci(int n) { |
| 105 | + if(n == 0) |
| 106 | + return 0; |
| 107 | + else if(n == 1) |
| 108 | + return 1; |
| 109 | + else |
| 110 | + return helpFibonacci(n -1 ) + helpFibonacci(n - 2); |
| 111 | + } |
| 112 | + /** |
| 113 | + * 返回小于给定最大值max的所有素数数组 |
| 114 | + * 例如max = 23, 返回的数组为[2,3,5,7,11,13,17,19] |
| 115 | + * @param max |
| 116 | + * @return |
| 117 | + */ |
| 118 | + public int[] getPrimes(int max){ |
| 119 | + return null; |
| 120 | + } |
| 121 | + |
| 122 | + /** |
| 123 | + * 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 |
| 124 | + * 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 |
| 125 | + * @param max |
| 126 | + * @return |
| 127 | + */ |
| 128 | + public int[] getPerfectNumbers(int max){ |
| 129 | + return null; |
| 130 | + } |
| 131 | + |
| 132 | + /** |
| 133 | + * 用seperator 把数组 array给连接起来 |
| 134 | + * 例如array= [3,8,9], seperator = "-" |
| 135 | + * 则返回值为"3-8-9" |
| 136 | + * @param array |
| 137 | + * @param |
| 138 | + * @return |
| 139 | + */ |
| 140 | + public String join(int[] array, String seperator){ |
| 141 | + return null; |
| 142 | + } |
| 143 | + |
| 144 | + public static void main(String[] args) { |
| 145 | + int[] a = fibonacci(15); |
| 146 | + for (int i = 0; i < a.length; i++) { |
| 147 | + System.out.println(a[i]); |
| 148 | + } |
| 149 | + } |
| 150 | +} |
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