leetcode-2-Add Two Numbers

题目：Add Two Numbers

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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.


思路：一开始觉得这题不难，不就是直接将两个链表的数都取出来求和，然后再构造一个新链表将每一位再插回去，写出如下代码

package leetcodeTest;
import java.util.*;
public class addTwoNumbersClass {
    public static void main(String[]args){
    int []a={9};
    int[]b={1,9,9,9,9,9,9,9,9,9};
    ListNode l1=creatList(a);
    ListNode l2=creatList(b);
      ListNode l=  addTwoNumbers( l1, l2);
      while(l!=null){
          System.out.println(l.val);
          l=l.next;
      }
    }

    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1==null&&l2==null) return null;
        int a=0;
        long sum1=l1.val,sum2=l2.val,sum=0;
        int m=0,n=0;//m,n代表L1,L2的长度
        int b=0;
        while(l1!=null){
            sum1+=a*(long)(Math.pow(10,m));
            l1=l1.next;
            m++;
            if(l1==null)break;
            a=l1.val;
        }
        while(l2!=null){
            sum2+=b*((long)(Math.pow(10,n)));
            l2=l2.next;
            n++;
            if(l2==null)break;
            b=l2.val;
        }
        ListNode s = new ListNode(0);
        ListNode r=s;
        sum=sum1+sum2;
        if(sum==0)return new ListNode(0);
        while(sum!=0) {
            int p = (int) (sum % 10);
            ListNode t=new ListNode(p);
            s.next = t;
            s = s.next;
            sum = sum / 10;
        }
        return r.next;
    }

public static ListNode creatList(int[] x){
        //创建链表
    ListNode q = new ListNode(0);
    ListNode w=q;
    for(int i=0;i<x.length;i++) {
        ListNode p = new ListNode(x[i]);
        q.next = p;
        q=q.next;
    }
    return  w.next;
}




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

然而我还是太年轻了，就算用上了long整形，测试数据的链表长度直接达到了60多位后，如果直接转化成数求和的话，long 也不够用这就说明思路不对

再仔细观察一下，发现其就是一个简单的向右进位问题，取两链表的当前位直接相加，若无进位就直接将其放入新链表节点中，若有进位则将需要进位的数加入到下一位相加的数中，然后取10的余放入当前位的链表节点中，然后就行了。Your runtime beats 87.03 % of java submissions

package leetcodeTest;
import java.util.*;
public class addTwoNumbersClass {
    public static void main(String[]args){
    int []a={2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,9};
    int[]b={5,6,4,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,9,9,9,9};

    ListNode l1=creatList(a);
    ListNode l2=creatList(b);
      ListNode l=  addTwoNumbers( l1, l2);
      while(l!=null){
          System.out.println(l.val);
          l=l.next;
      }
    }

    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1==null&&l2==null) return null;
        ListNode s = new ListNode(0);
        ListNode r=s;
        int m=0;
        while(l1!=null&&l2!=null){
            int a=l1.val;
            int b=l2.val;
            int sum=a+b+m;
            m=sum/10;//向右进位的数
            int n=sum%10;//当前位
            ListNode t=new ListNode(n);
            s.next=t;
            l1=l1.next;
            l2=l2.next;
            s=s.next;
        }
        while(l1==null&&l2!=null){
            int b=l2.val;
            int sum=b+m;
            m=sum/10;
            int n=sum%10;
            ListNode t=new ListNode(n);
            s.next=t;
            l2=l2.next;
            s=s.next;
        }
        while(l1!=null&&l2==null){
            int a=l1.val;
            int sum=a+m;
            m=sum/10;
            int n=sum%10;
            ListNode t=new ListNode(n);
            s.next=t;
            l1=l1.next;
            s=s.next;
        }
        if(m!=0){
           //若l1,l2都为空时还有进位，则需要将该进位新加入链表中
            ListNode t=new ListNode(m);
            s.next=t;
        }
        return r.next;
    }

public static ListNode creatList(int[] x){
        //创建链表
    ListNode q = new ListNode(0);
    ListNode w=q;
    for(int i=0;i<x.length;i++) {
        ListNode p = new ListNode(x[i]);
        q.next = p;
        q=q.next;
    }
    return  w.next;
}




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

 ac后，学习了一下排名前面的代码，可以看这个代码，它通过将l1或l2为null后将其值直接赋0，免去了我的代码当中对l1==null和l2==null时的分开讨论，写的很简洁，很巧妙！

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode l3 = new ListNode(0);
        ListNode p = l1, q = l2, current = l3;
        int carry = 0;
        while (p != null || q != null) {
            int x = (p == null) ? 0 : p.val;
            int y = (q == null) ? 0 : q.val;
            int sum = x + y + carry;
            carry = sum/10;
            current.next = new ListNode(sum%10);
            current = current.next;
            if(p!=null) p = p.next;
            if(q!=null) q = q.next;
        }
        if (carry > 0) {
            current.next = new ListNode(carry);
            current = current.next;
        }
        return l3.next;
    }
}