用JAVA写一个函数。功能如下:任意给定一组数,例如{12,60,-8,99,15,35,17,18},找出任意数相加之后的结果为35(任意设定)的情况.
可以递归算法来解:
package test1; import java.util.Arrays; public class demo { public static void main(String[] args) { String str = "12,60,-8,99,15,35,17,18,8,10,11,12"; int sum = 35; diguiSum(str,sum); } public static void diguiSum(String str,int sum) { String[] x = str.split(","); int[] array = arrayTransform(x); for (int i = 0; i < array.length; i++) { int[] cache = new int[i + 1]; int ceng = -1; int cengQuit = i; int startPiont = 0; cir(ceng, cengQuit, startPiont, array, cache, sum); } } // 递归求结果 public static void cir(int ceng, int cengQuit, int startPiont, int[] array, int[] cache, int sum) { ceng++; for (int i = startPiont; i < array.length; i++) { cache[ceng] = array[i]; if (ceng == cengQuit) { if (getSum(cache) == sum) { printcache(cache); } if (getSum(cache) > sum) { break; } } if (ceng < cengQuit) { startPiont = i + 1; cir(ceng, cengQuit, startPiont, array, cache,sum); } } } // 获取组合数字之和 public static int getSum(int[] cache) { int sum = 0; for (int i = 0; i < cache.length; i++) { sum = sum + cache[i]; } return sum; } // 打印组合的可能 public static void printcache(int[] cache) { for (int i = 0; i < cache.length; i++) { System.out.print(cache[i] + ","); } System.out.println(); } // 转换数组类型 且为提高效率做准备 public static int[] arrayTransform(String[] strArray) { int length = 0; int[] array = new int[strArray.length]; for (int i = 0; i < strArray.length; i++) { array[i] = Integer.valueOf(strArray[i]); } Arrays.sort(array); for (int i = 0; i < array.length; i++) { if (array[i] > 35) { length = i; break; } } int[] dest = new int[length]; System.arraycopy(array, 0, dest, 0, length); return dest; } }此算法有做适当参考。
运行结果:
用JAVA写一个函数,功能如下: 任意给定一组数, 找出任意数相加之后的结果为35(任意设定)的情况,布布扣,bubuko.com
用JAVA写一个函数,功能如下: 任意给定一组数, 找出任意数相加之后的结果为35(任意设定)的情况
原文地址:http://blog.csdn.net/chickencin/article/details/28425183