Arithmetic<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />
Please complete the function according to the document demand
public class Ncsmsoft {
public static void main(String[] args) {
//test result
System.out.print(findSum(35,"12,60,8,-8,99,15,35,17,18"));
}
/*==============================================
function：findSum
creator：(name)
create time：(time)
updator name：(name)
update time：(time)
version number：1.0
description：find all the possible assembled from the character string that the sum of which
equal to the appoint value
input parameter：Double he //appoint value
String shuzu //a character string divide by comma
output parameter：result
output format：for example "12+8+15;35;17+18"
testor ：(name)
=================================================*/
private static String findSum(double he,String shuzu){
StringBuffer result=new StringBuffer();
//Please input your program here without changing of the other code
//return your result as the format "12+8+15;35;17+18"
return result.toString();
}
}
answer:
package com.softeem.demo;
import java.util.ArrayList;
import java.util.Arrays;
public class Ncsmsoft {
public static void main(String[] args) {
// test result
System.out.print(findSum(35, "12,60,8,-8,99,15,35,17,18"));
}
/*
* ============================================== function：findSum
* creator：(name) create time：(time) updator name：(name) update time：(time)
* version number：1.0 description：find all the possible assembled from the
* character string that the sum of which equal to the appoint value input
* parameter：Double he //appoint value String shuzu //a character string
* divide by comma output parameter：result output format：for example
* "12+8+15;35;17+18" testor ：(name)
* =================================================
*/
private static String findSum(double he, String shuzu) {
StringBuffer result = new StringBuffer();
// Please input your program here without changing of the other code
String[] strs = shuzu.split(",");
int[] a = new int[strs.length];
for (int i = 0; i < a.length; i++) {
a[i] = Integer.parseInt(strs[i]);
}
Arrays.sort(a); // 对初始数据进行排序，便于算法的实现
ArrayList<Integer> okList = new ArrayList<Integer>(); // 存放的是当前找到的合法的元素，这些元素的和小于SUM
find(a, he, 0, okList, result); // 通过递归的方式进行查找
// return your result as the format "12+8+15;35;17+18"
return result.toString();
}
/**
* 递归函数，每一次的动作很简单,在已经找到的n个元素的基础上,寻找第n+1个元素
*/
private static void find(int[] a, final double SUM, int cur,
ArrayList<Integer> okList, StringBuffer result) {
int beg = okList.size() == 0 ? 0 : okList.get(okList.size() - 1) + 1; // 当前元素的查找范围的起始位置
for (int i = beg; i < a.length; i++) { // 从起始位置到结束位置，查找合适的元素
cur += a[i]; // 在前面元素的和的基础上，加上当前元素
if (cur < SUM) { // 如果仍然小于SUM，证明当前元素(第n+1个)合法，继续寻找第n+2个元素
okList.add(i);
find(a, SUM, cur, okList, result);
cur -= a[i]; // 消除第i个元素的影响，为了试验第i+1个元素做准备
okList.remove(okList.size() - 1); // 消除第i个元素的影响，为了试验第i+1个元素做准备
} else if (cur == SUM) { // 如果等于SUM，证明找到了表达式
for (int x : okList) { // 构造表达式，并存入result
result.append(a[x]);
result.append("+");
}
result.append(a[i]);
result.append(";");
break; // 回溯到上一个状态
} else { // 如果大于SUM，也回溯到上一个状态
break;
}
}
}
}