[leetcode]2.add two numbers

突然碰到一个medium的题，是真的medium啊，一下子就越界了，先没有用脑子用的int，然后发现long也总是转化成int，遂一怒之下用了biginteger，然后

Success
Details
Runtime: 173 ms, faster than 0.98% of Java online submissions for Add Two Numbers.

我哭了
我这个脑子

知识点

• java的单链表就是如此简洁，清晰，就在对象里面放对象，相比c的指针来说
• 构造函数如果重载了默认的，那么一定要用这个重载的构造进行初始化。比如这个题中， ListNode l3=new ListNode(0);一定要传参0，不能为空
• java所有变量都要赋初始值
• 对象如果只是为了和其他对象指向同一个空间，那可以不new，只声明类型
  像ListNode p;
  p=l1;
  这样p的val next实际都是l1的，不用再自己构造
  但如果这个p是想是要放进链表的结点，那么要new出来整个结构

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
import java.math.*;
class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode p;
        p=l1;
        BigInteger k=new BigInteger("1");
        BigInteger num1=new BigInteger("0");
        BigInteger num2=new BigInteger("0");
         BigInteger temp1=new BigInteger("0");
         BigInteger temp2=new BigInteger("0");
        while(p!=null){
            temp1=k.multiply(BigInteger.valueOf(p.val));
            num1=num1.add(temp1);
            System.out.println(num1);
            k=k.multiply(new BigInteger("10"));
            p=p.next;
        }
        p=l2;
        k=new BigInteger("1");
        while(p!=null){
            temp2=k.multiply(BigInteger.valueOf(p.val));
            num2=num2.add(temp2);
            System.out.println(num2);
            k=k.multiply(new BigInteger("10"));
            p=p.next;
        }
        ListNode l3=new ListNode(0);
        p=l3;
        BigInteger sum=new BigInteger("0");
        sum=num1.add(num2);
        System.out.println(sum);
        while(sum.compareTo(new BigInteger("0"))>0){
            p.val=(sum.remainder(new BigInteger("10"))).intValue();
            sum=sum.divide(new BigInteger("10"));
            if(sum.compareTo(new BigInteger("0"))!=0){
                ListNode q=new ListNode(0);
                q.next=null;
                p.next=q;
                p=q; 
            }
                
        }
        p.next=null;
        return l3;
        
    }
}

后来仿造思路写了一遍

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode p,q,curr;
        p=l1;
        q=l2;
        ListNode l3=new ListNode(0);
        curr=l3;
        int i=0;
        int x=0,y=0;
        while(p!=null||q!=null){
            if(p==null){
                x=0;
            }else{
                x=p.val;
                p=p.next;
            }
            if(q==null){
                y=0;
            }else{
                y=q.val;
                q=q.next;
            }
            int sum=x+y+i;
            if(sum>=10)
            {
                i=1;
            }
            else
            {
                i=0;
            }
            curr.val=sum%10;
            curr.next=null;
            
            if(p!=null||q!=null){
            ListNode n=new ListNode(0);
            curr.next=n;
            curr=n;
            }
           
        }
        if(i==1){
            ListNode a=new ListNode(1);
            curr.next=a;
            a.next=null;
        }
        return l3;
    }
}

标准答案用了头结点，这样就不用管最后一个节点了

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}