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网上看到的5个问题，下面是解答，不知道有没有其他建议！
问题1
使用for循环、while循环和递归写出3个函数来计算给定数列的总和。
package com.luka; public class Javat { private static int[] arr_Ints = { 2, 1, 4, 3, 6, 5, 8, 7, 10, 9 }; public static void main(String[] args) { System.out.println("The Count is " + getNumByFor() + " ."); System.out.println("The Count is " + getNumByWhile() + " ."); System.out.println("The Count is " + getNumByEcursion(0) + " ."); } /** * for 循环 */ private static int getNumByFor() { int count = 0; for (int i = 0; i < arr_Ints.length; i++) { count += arr_Ints[i]; } return count; } /** * while 循环 */ private static int getNumByWhile() { int count = 0; int i = 0; while (i < arr_Ints.length) { count += arr_Ints[i]; i++; } return count; } /** * 递归 */ private static int getNumByEcursion(int i) { if (arr_Ints.length == 0) return 0; else if (i < arr_Ints.length - 1) return arr_Ints[i] + getNumByEcursion(i + 1); else return arr_Ints[i]; } }
问题2
编写一个交错合并列表元素的函数。例如：给定的两个列表为[a，B，C]和[1，2，3]，函数返回[a，1，B，2，C，3]。
package com.luka; public class Javat { private static String[] arr1 = { "a", "B", "c", "D", "e" }; private static String[] arr2 = { "1", "2", "3" }; public static void main(String[] args) { String[] arr = getNum(arr1, arr2); for (int i = 0; i < arr.length; i++) System.out.println("The Num is " + arr[i] + " ."); } private static String[] getNum(String[] arr12, String[] arr22) { String[] arr = new String[arr1.length + arr2.length]; int i, j; for (i = 0, j = 0; i < arr1.length; i++) { j = 2 * i; if (j > 2 * arr2.length) j = arr2.length + i; arr[j] = arr1[i]; } for (i = 0, j = 0; i < arr2.length; i++) { j = 2 * i + 1; if (j > 2 * arr1.length) j = arr1.length + i; arr[j] = arr2[i]; } return arr; } }
问题3
编写一个计算前100位斐波那契数的函数。根据定义，斐波那契序列的前两位数字是0和1，随后的每个数字是前两个数字的和。例如，前10位斐波那契数为：0，1，1，2，3，5，8，13，21，34。
package com.luka; public class Javat { public static void main(String[] args) { try { System.out.println("The Nums is " + getCount(100) + " ."); } catch (Exception e) { } } // 获取值 private static int getNum(int num) { int count = 0; if (num <= 1) count = 0; else if (num == 2) count = 1; else count = getNum(num - 1) + getNum(num - 2); return count; } // 获取和 private static String getCount(int num) { String strNums = ""; for (int i = 0; i <= num; i++) { strNums += getNum(i) + ","; } strNums = strNums.substring(0, strNums.length()-1); return strNums; } }
问题4
编写一个能将给定非负整数列表中的数字排列成最大数字的函数。例如，给定[50，2，1,9]，最大数字为95021。
package com.luka; import java.util.Arrays; import java.util.Comparator; public class Javat { private static Integer[] VALUES = { 50, 2, 100, 99, 5, 7, 51,50 ,11}; public static void main(String[] args) { Arrays.sort(VALUES, new Comparator<Object>() { @Override public int compare(Object lhs, Object rhs) { String v1 = lhs.toString(); String v2 = rhs.toString(); return (v1 + v2).compareTo(v2 + v1) * -1; } }); String result = ""; for (Integer integer : VALUES) { result += integer.toString(); } System.out.println(result); } }
问题5
编写一个在1，2，…，9（顺序不能变）数字之间插入+或-或什么都不插入，使得计算结果总是100的程序，并输出所有的可能性。例如：1 + 2 + 34 – 5 + 67 – 8 + 9 = 100。
package com.luka; import java.util.ArrayList; public class Javat { private static int TARGET_SUM = 100; private static int[] VALUES = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; private static ArrayList add(int digit, String sign, ArrayList branches) { for (int i = 0; i < branches.size(); i++) { branches.set(i, digit + sign + branches.get(i)); } return branches; } private static ArrayList f(int sum, int number, int index) { int digit = Math.abs(number % 10); if (index >= VALUES.length) { if (sum == number) { ArrayList result = new ArrayList(); result.add(Integer.toString(digit)); return result; } else { return new ArrayList(); } } ArrayList branch1 = f(sum - number, VALUES[index], index + 1); ArrayList branch2 = f(sum - number, -VALUES[index], index + 1); int concatenatedNumber = number >= 0 ? 10 * number + VALUES[index] : 10 * number - VALUES[index]; ArrayList branch3 = f(sum, concatenatedNumber, index + 1); ArrayList results = new ArrayList(); results.addAll(add(digit, "+", branch1)); results.addAll(add(digit, "-", branch2)); results.addAll(add(digit, "", branch3)); return results; } public static void main(String[] args) { ArrayList list = f(TARGET_SUM, VALUES[0], 1); for(int i=0;i<list.size();i++) { System.out.println(list.get(i)); } } }
输出结果：
1+2+3-4+5+6+78+9 1+2+34-5+67-8+9 1+23-4+5+6+78-9 1+23-4+56+7+8+9 12+3+4+5-6-7+89 12+3-4+5+67+8+9 12-3-4+5-6+7+89 123+4-5+67-89 123+45-67+8-9 123-4-5-6-7+8-9 123-45-67+89
来源：http://developer.51cto.com/art/201505/476097.htm
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原文地址：http://www.cnblogs.com/yc-755909659/p/4501245.html
