java合并2个链表

Question

我正在尝试将两个链表合并在一起，列表1和列表2。在列表2中，您有一个在列表1中不可用的项目，即键盘，同时鼠标数量也已更改，新的鼠标数量将为3，因为在列表1中有1个鼠标数量，而在列表2中有3个。所以本质上你会有一个新的链表，叫做list3。list3将list1和list2的内容组合在一起，就像list2中的项目键盘，而不是list1，还有新的鼠标数量

import java.util.LinkedList; import java.util.ListIterator; public class TestLinkedList extends MyLinkedList { public static void main(String[] args) { LinkedList < String > list1 = new LinkedList < > (); Demo demo = new Demo(); String case1 = demo.setNameandQty(1, "Case"); String monitor1 = demo.setNameandQty(3, "Monitor"); String mouse1 = demo.setNameandQty(1, "Mouse"); String ram1 = demo.setNameandQty(2, "RAM"); String ssd1 = demo.setNameandQty(4, "SSD"); int cm = 2; list1.add(case1); list1.add(monitor1); list1.add(mouse1); list1.add(ram1); list1.add(ssd1); System.out.println("Shopping List 1"); ListIterator < String > it1 = list1.listIterator(); while (it1.hasNext()) { System.out.println(it1.next()); } list1.remove(3); System.out.println("\nAfter deleting 4th item"); for (String s: list1) { System.out.println(s); } LinkedList < String > list2 = new LinkedList < > (); String keyboard2 = demo.setNameandQty(1, "Keyboard"); String mouse2 = demo.setNameandQty(2, "Mouse"); list2.add(keyboard2); list2.add(mouse2); System.out.println("\nShopping list 2"); for (String s: list2) { System.out.println(s); } String holdMouse1q = ""; String holdMouse2q = ""; String getMouse1q = demo.getIntegers(mouse1, holdMouse1q); int mouse1q = Integer.parseInt(getMouse1q); String getMouse2q = demo.getIntegers(mouse2, holdMouse2q); int mouse2q = Integer.parseInt(getMouse2q); System.out.println("\nMouse 1 quantity: " + mouse1q); System.out.println("Mouse 2 quantity: " + mouse2q); int totalMouseQ = mouse1q + mouse2q; // list1.addFirst("Flashdisks : 10"); // System.out.println("\nAfter adding 10 flash disks"); // for (String s : list1) { // System.out.println(s); // } } }

Answer 1

我不确定方法demo.setNameandQty()的返回值是什么。所以我假设返回值应该像<name>-<quality>，例如“Mouse-2”

下面是我的合并方法，通过对项目列表排序进行求解：

private static List<String> merge(List<String> list1, List<String> list2){
Collections.sort(list1);
Collections.sort(list2);
//ensure list always has 1 element
list1.add(null);
list2.add(null);

List<String> list3 = new LinkedList<String>();

Iterator<String> l1 = list1.iterator();
Iterator<String> l2 = list2.iterator();

String item1 = l1.next();
String item2 = l2.next();
while(item1 != null && item2 != null){
    String name1 = getName(item1);
    String name2 = getName(item2);
    if(name1.compareTo(name2) < 0){
        list3.add(item1);
        item1 = l1.next();
    }
    else if(name1.compareTo(name2) > 0){
        list3.add(item2);
        item2 = l2.next();
    }
    else if(name1.compareTo(name2) == 0){
        //sum up the quality then add to the list
        int totalQty = getQuality(item1) + getQuality(item2);
        list3.add(name1 + "-" + totalQty);
        item1 = l1.next();
        item2 = l2.next();
    }
}

//add the rest of the list, in case one of two lists still has items
while(item1 != null){
    list3.add(item1);
    item1 = l2.next();
}
while(item2 != null){
    list3.add(item2);
    item2 = l2.next();
}

return list3;
}    
private static String getName(String s){
    return s.substring(0, s.indexOf('-'));
}
private static int getQuality(String s){
    return Integer.valueOf(s.substring(s.indexOf('-')+1));
}

这里需要注意的是，如果您在一个列表中有重复的项，例如{“Mouse-1”、“Mouse-2”}，则此方法将失败。为了解决这个问题，我认为我们最好确保没有重复项，而不是使合并方法更加复杂