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;
	}
}

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