如何在我的java代碼中刪除這些嵌套的for-loops？

Question

我在算法類中，我們需要製作一個程序，通過將它們加載到鏈表中來添加很長的數字。我在網上看到很多例子，只使用兩個鏈表，但我的需求不止兩個。

import java.util.*; 


public class longNumbersLinkedListCompleted 
{ 

public static void main(String[] args) 
{ 
    Scanner stdIn = new Scanner(System.in); 
    String longNumber = ""; 
    LinkedList mainList = new LinkedList(); 
    LinkedList sumList = new LinkedList(); 
    LinkedList temp = null; 

    //declare other variables 
    int sum = 0; 
    int carry= 0; 
    int maxWidth = 0; 

    System.out.println("Enter a Number"); 
    longNumber = stdIn.nextLine(); 
    //repeatedly input longNumbers, using -1 to indicate that you are done 
    while(!longNumber.equals("-1")){ 

     //add a new LinkedList at the beginning of the mainList 
     mainList.addFirst(new LinkedList()); 

     //use get(0) to set temp to be this new LinkedList 
     temp = (LinkedList) mainList.get(0); 

     //for each character in your longNumber, subtract 48 to get the digit and then add it 
     //at the beginning of temp 
     for (int i = 0; i < longNumber.length(); i++){ 
      temp.addFirst((int)(longNumber.charAt(i)-48));   
     } 

     //keep track of maxWidth, the number of digits of the widest longNumber input so far 
     if(maxWidth < longNumber.length()){ 
      maxWidth = longNumber.length(); 
     } 

     System.out.println("Enter a Number"); 
     longNumber = stdIn.nextLine(); 
    } 

    //make maxWidth passes 
    //initialize carry to be 0 
    //in each pass, loop through all of the LinkedLists in mainList 
    //for each one, let temp be the Linked List for one longNumber 
    //if temp is not empty, remove its first entry and add to the sum 
    for(int i = 0; i < maxWidth; i++){ 
     sum = 0; 

     for(int j = 0; j < mainList.size(); j++){ 
      temp = (LinkedList)mainList.get(j); 
      if(temp.size() > 0){ 
       sum += (int)temp.removeFirst(); 
      } 
     } 
     //add sum%10 at the beginning of sumList 
     //set carry equal to sum/10 (integer division) 
     sumList.addFirst((sum+carry)%10); 
     carry = (sum + carry)/10; 
    } 

    //Now ready for output 
    //if carry at the end of processing is not 0, print it and stay on the same line 
    //repeatedly remove one digit from the beginning of sumList until all have been removed 
    //for each, add a 48 to get a character and print it out on the same line 
    if(carry != 0){ 
     System.out.print(carry); 
    } 
    for(int i = sumList.size(); i > 0; i--){ 
     System.out.print((char)((int)sumList.removeFirst()+48)); 
    } 

    //remove the digits from sumList until empty 

}//end main 

} //結束類

的意見是從教授，所以它有點聽起來像他想的嵌套的循環，但在當今一流他提到不使用他們，因爲整個類是關於使代碼更好，使用O（n^2）的東西似乎是不直觀的。

我很感謝你們的幫助！

Answer 1

我可以看到避免嵌套for循環的唯一方法是在讀取長整數時進行轉置，即在您正在處理時在while循環中構建和數。

喜歡的東西：

List<Integer> sums = new ArrayList<>(); 
while(!longNumber.equals("-1")){ 
    for (int i = 0; i < longNumber.length(); ++i) { 
    // Process the digits in reverse. 
    int digit = Character.getNumericValue(longNumber.charAt(longNumber.length() - 1 - i)); 

    if (i >= sums.length()) { 
     sums.add(digit); 
    } else { 
     sums.set(i, sums.get(i) + digit); 
    } 
    } 

    longNumber = stdIn.nextLine(); 
} 

（請注意，這仍然是一個嵌套循環，你只需要做的是進行循環，直至您已經閱讀所有的號碼，並閱讀這些數字每一個數字） 。

然後將以下內嵌套的for循環是不必要的，因爲總和已經計算：

// This is the equivalent of the for(int i = 0; i < maxWidth; i++){ loop. 
for (int sum : sums) { 
    sumList.addFirst((sum+carry)%10); 
    carry = (sum + carry)/10; 
} 

（當然，也可以摺疊該第二環路進入while循環，並避免在製造sums所有）。