Java:数组的可能组合
Java:数组的可能组合
提问于 2018-09-30 07:06:10
我正在解决一个java问题：
问题：
用户输入一个整数数组，该数组的大小可以是1-9。我需要重新排列数组，以便它打印出可以被3整除的最大值。我不需要使用数组中的所有整数。
示例：
输入：(int list) l= 3，1，4，1
输出：(int) 4311
输入：(int list) l= 3，1，4，1，5，9
输出：(int) 94311
--
到目前为止，我在这上面花了大约5个小时。我的代码可以工作，但它几乎尝试了每种组合，直到其中一种工作。我需要一个更有效的代码。
这是我的代码：
public class CodedMessage {
public static int zero = 0;
public static int one = 0;
public static int two = 0;
public static int three = 0;
public static int four = 0;
public static int five = 0;
public static int six = 0;
public static int seven = 0;
public static int eight = 0;
public static int nine = 0;
public static int best = 0;
public static int current = 0;
public static int newZero = 0;
public static int newOne = 0;
public static int newTwo = 0;
public static int newThree = 0;
public static int newFour = 0;
public static int newFive = 0;
public static int newSix = 0;
public static int newSeven = 0;
public static int newEight = 0;
public static int newNine = 0;
public static void main(String[] args) {
int[] myIntArray = {3,1,4,1};
System.out.println(answer(myIntArray));
}
public static int answer(int[] l) {
String line ="";
for (int c : l) {
line += c;
}
zero = line.length() - line.replace("0", "").length();
one = line.length() - line.replace("1", "").length();
;
two = line.length() - line.replace("2", "").length();
;
three = line.length() - line.replace("3", "").length();
;
four = line.length() - line.replace("4", "").length();
;
five = line.length() - line.replace("5", "").length();
;
six = line.length() - line.replace("6", "").length();
;
seven = line.length() - line.replace("7", "").length();
;
eight = line.length() - line.replace("8", "").length();
;
nine = line.length() - line.replace("9", "").length();
;
if (Integer.parseInt(line)%3 != 0) {
}else {
possibleStrings(l.length, l, "");
}
return best;
}
public static String possibleStrings(int maxLength, int[] alphabet, String curr) {
if (!curr.equals("")) {
current = Integer.parseInt(curr);
}
if (current > best) {
if (current % 3 == 0) {
String line = Integer.toString(current);
newZero = line.length() - line.replace("0", "").length();
newOne = line.length() - line.replace("1", "").length();
;
newTwo = line.length() - line.replace("2", "").length();
;
newThree = line.length() - line.replace("3", "").length();
;
newFour = line.length() - line.replace("4", "").length();
;
newFive = line.length() - line.replace("5", "").length();
;
newSix = line.length() - line.replace("6", "").length();
;
newSeven = line.length() - line.replace("7", "").length();
;
newEight = line.length() - line.replace("8", "").length();
;
newNine = line.length() - line.replace("9", "").length();
;
if (zero >= newZero && one >= newOne && two >= newTwo && three >= newThree && four >= newFour
&& five >= newFive && six >= newSix && seven >= newSeven && eight >= newEight
&& nine >= newNine) {
best = current;
}
}
}
if (curr.length() == maxLength) {
if (!curr.equals("") &&Integer.parseInt(curr)%3 != 0) {
return "hi";
}
} else {
for (int i = 0; i < alphabet.length; i++) {
String oldCurr = curr;
curr += alphabet[i];
possibleStrings(maxLength, alphabet, curr);
curr = oldCurr;
}
}
return "hi";
}
}
有人能让它变得更有效率吗?我试过了，但没能做到。
谢谢!
回答 1
Stack Overflow用户
发布于 2018-09-30 09:27:43
您只需要按数字的降序对数组进行排序，并将它们写在一行中。这能解决问题吗？
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/52573261
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