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  • 【Leetcode】2. Add Two Numbers

    Problem:

    You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

    You may assume the two numbers do not contain any leading zero, except the number 0 itself.

    Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
    Output: 7 -> 0 -> 8

     (分析:相同长度的链表对应val相加,大于10的本位val为0,.next.val加1)

    Resolution

    public ListNode addTwoNumbers(ListNode l1,ListNode l2){
        int carry =0;//每一位的数值
        ListNode p,result = new ListNode(0);
        p=result;
        while(l1!=null||l2!=null||carry!=0){//注意是l1与l2不为空,而不是l1.next与l2.next不为空
            if(l1!=null){
                carry+=l1.val;
                l1=l1.next;
            }
            if (l2!=null){
               carry+=l2.val;
               l2=l2.next;
            }
            p.next=new ListNode(carry%10);
            carry/10;
            p=p.next;
        }
        return result.next;
    }
    

     注:

    ①需要保证l1或l2非空。在carry!=0时也要继续进行计算的原因是当l1与l2均进行完加法,到达链表末尾,不会出现溢出问题。

    ②处理方式并非相对l1与l2的val加和在赋值给carry,而是carry+=l1.val,在carry+=l2.val;

    ③while循环时,注意是l1与l2不为空,而不是l1.next与l2.next不为空

    ④ListNode p,result = new ListNode(0); p=result; 初始化的顺序为p,result 后面赋值也必须是这个顺序,否则会报错。

    完整可运行代码如下:

    package medium;
    
    /**
     * Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
    	Output: 7 -> 0 -> 8
     * @author yum717
     *
     */
    public class L2 {
    	public static class ListNode {
    		int val;
    		ListNode next;
    		ListNode(int x) {
    			val = x;
    		}
    	}
    
    	public  ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    	    int carry = 0;
    	    ListNode p, result = new ListNode(0);
    	    p = result;
    	    while (l1 != null || l2 != null || carry != 0) {
    	        if (l1 != null) {
    	            carry += l1.val;
    	            l1 = l1.next;
    	        }
    	        if (l2 != null) {
    	            carry += l2.val;
    	            l2 = l2.next;
    	        }
    	        p.next = new ListNode(carry%10);
    	        carry /= 10;
    	        p = p.next;
    	       
    	    } 
            //输出测试 while(result.next!=null){ System.out.println(result.next.val); result=result.next;   } return result.next; } public static void main(String[] args) { ListNode l1 = new ListNode(2); ListNode l2= new ListNode(5); ListNode l3 = new ListNode(4); ListNode l4= new ListNode(3); l1.next=l3; l3.next=l4; l4.next=null; ListNode l23= new ListNode(6); ListNode l24 = new ListNode(4); l2.next=l23; l23.next=l24; l4.next=null; L2 l= new L2(); l.addTwoNumbers(l1,l2); } }

     问题:1.实例化一个链表还有什么简洁的的办法?

    2.public static class ListNode { int val; ListNode next; ListNode(int x) { val = x; } }

    定义链表时,如果不加入static 会出现如下报错信息,不知道如何解释。

    Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
    	No enclosing instance of type L2 is accessible. Must qualify the allocation with an enclosing instance of type L2
         (e.g. x.new A() where x is an instance of L2).
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  • 原文地址:https://www.cnblogs.com/yumiaomiao/p/7029202.html
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