交替合並兩個鏈接列表，同時尋找最大的頭部

Question

我正在開發一個鏈表程序，該程序只能循環遍歷該列表一次，而且無法將列表的元素復制到另一個數據結構中。

假設列表不為空（至少有一個節點），而最后一個節點的下一個為空。

下面的方法應該合並兩個鏈表，合並表的頭應該是兩個鏈表中的最大值。 以下值在兩個列表之間交替顯示。

例如，如果我的輸入是：

1 2 4 

3 5 6 7 8 9

我的輸出將是：

3 1 5 2 6 4 7 8 9

我想出了如何分配頭的方法，但是我想不出如何正確地合並他的其余部分。

這是我的代碼：

    public class LinkedListNode {
    private int value;
    private LinkedListNode next;

    public LinkedListNode(int value) {
        this.value = value;
    }

    public int getValue() {
        return value;
    }

    public void setValue(int value) {
        this.value = value;
    }

    public LinkedListNode getNext() {
        return next;
    }

    public void setNext(LinkedListNode next) {
        this.next = next;
    }
}

    public static LinkedListNode mergeLists (LinkedListNode head1, LinkedListNode head2){
        LinkedListNode firstList = head1;
        LinkedListNode secondList = head2;
        LinkedListNode tmp = new LinkedListNode(0);
        LinkedListNode result = head1;
        int checkWhichList=0;
        while(firstList!=null && secondList!=null){
            if (firstList.getValue()<=secondList.getValue()){
                result=secondList;
            }
            if (checkWhichList%2==0){
                tmp.setNext(head2.getNext());
                checkWhichList++;
            }
            else
                tmp.setNext(head1.getNext());
        }

        result.setNext(tmp);
        return result;
    }

Answer 1

一旦確定哪個節點必須添加兩個頭最大值result （第一最大值），然后將您的指針firstList和secondList

現在，您繼續從兩個列表中添加結果，同時移動到每個列表中的下一個節點，直到兩個都沒有指向任何對象為止

示例：（以firstList為頭的情況為例）

if(firstList.getValue() > secondList.getValue()) {
    //here firstList will be the head
    result = firstList;
    result.setNext(secondList);
    firstList = firstList.getNext();
    secondList = secondList.getNext();

    //Now alternating (adding firstList node before secondList)
    while(firstList != null || secondList != null) {
        if(firstList != null) {
            result.setNext(firstList);
            firstList = firstList.getNext();
        }        
        if(secondList != null) {
            result.setNext(secondList);
            secondList = secondList.getNext();
        }
    }
} else {
   //here secondList will be the head
   //continue with the code
}

Answer 2

您可以嘗試以下代碼：

public class Test {
public static void main(String args[]) {
    //1 2 4
    //3 5 6 7 8 9
    LinkedListNode list1 = new LinkedListNode(1);
    list1.insert(list1, 2);
    list1.insert(list1, 4);
    LinkedListNode list2 = new LinkedListNode(3);
    list2.insert(list2, 5);
    list2.insert(list2, 6);
    list2.insert(list2, 7);
    list2.insert(list2, 8);
    list2.insert(list2, 9);
    LinkedListNode result = mergeLists(list1, list2);
    while(result!=null){
        System.out.print(result.getValue() + " ");
        result = result.getNext();
    }

}

public static LinkedListNode mergeLists (LinkedListNode head1, LinkedListNode head2){
    LinkedListNode firstList = head1;
    LinkedListNode secondList = head2;
    LinkedListNode tmp = new LinkedListNode(0);
    LinkedListNode result = new LinkedListNode(0);
    int checkWhichList=0;
    if(firstList!=null && secondList!=null){
        if (firstList.getValue()<=secondList.getValue()){
            result.insert(result, secondList.getValue());
            secondList = secondList.getNext();
            checkWhichList++;
        }else{
            result.insert(result, firstList.getValue());
            firstList = firstList.getNext();
            //checkWhichList++;
        }
    }
    while(firstList!=null && secondList!=null) {
        if (checkWhichList % 2 != 0) {
            result.insert(result, firstList.getValue());
            firstList = firstList.getNext();
            checkWhichList++;
        } else{
            result.insert(result, secondList.getValue());
            secondList = secondList.getNext();
            checkWhichList++;
        }
    }
    if(firstList!=null){
        while(firstList!=null) {
            result.insert(result, firstList.getValue());
            firstList = firstList.getNext();
        }
    }
    if(secondList!=null){
        while(secondList!=null) {
            result.insert(result, secondList.getValue());
            secondList = secondList.getNext();
        }
    }

    //result.setNext(tmp);
    return result.getNext();
}
}

class LinkedListNode {
    private int value;
    private LinkedListNode next;

    public LinkedListNode(int value) {
        this.value = value;
    }

    public int getValue() {
        return value;
    }

    public void setValue(int value) {
        this.value = value;
    }

    public LinkedListNode getNext() {
        return next;
    }

    public void setNext(LinkedListNode next) {
        this.next = next;
    }

    public LinkedListNode insert(LinkedListNode head, int value){
        LinkedListNode temp = new LinkedListNode(value);
        LinkedListNode temp1 = head;
        if(head == null){
            return temp;
        }
        else{
            temp1 = head;
            while(temp1.next != null){
                temp1 = temp1.next;
            }
            temp1.next = temp;
        }
        return head;
    }
}