标签：AC 学习 测试 link 简单 ring bsp sub div
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
1 package leetcodeTest; 2 import java.util.*; 3 public class addTwoNumbersClass { 4 public static void main(String[]args){ 5 int []a={9}; 6 int[]b={1,9,9,9,9,9,9,9,9,9}; 7 ListNode l1=creatList(a); 8 ListNode l2=creatList(b); 9 ListNode l= addTwoNumbers( l1, l2); 10 while(l!=null){ 11 System.out.println(l.val); 12 l=l.next; 13 } 14 } 15 16 public static ListNode addTwoNumbers(ListNode l1, ListNode l2) { 17 if(l1==null&&l2==null) return null; 18 int a=0; 19 long sum1=l1.val,sum2=l2.val,sum=0; 20 int m=0,n=0;//m,n代表L1,L2的长度 21 int b=0; 22 while(l1!=null){ 23 sum1+=a*(long)(Math.pow(10,m)); 24 l1=l1.next; 25 m++; 26 if(l1==null)break; 27 a=l1.val; 28 } 29 while(l2!=null){ 30 sum2+=b*((long)(Math.pow(10,n))); 31 l2=l2.next; 32 n++; 33 if(l2==null)break; 34 b=l2.val; 35 } 36 ListNode s = new ListNode(0); 37 ListNode r=s; 38 sum=sum1+sum2; 39 if(sum==0)return new ListNode(0); 40 while(sum!=0) { 41 int p = (int) (sum % 10); 42 ListNode t=new ListNode(p); 43 s.next = t; 44 s = s.next; 45 sum = sum / 10; 46 } 47 return r.next; 48 } 49 50 public static ListNode creatList(int[] x){ 51 //创建链表 52 ListNode q = new ListNode(0); 53 ListNode w=q; 54 for(int i=0;i<x.length;i++) { 55 ListNode p = new ListNode(x[i]); 56 q.next = p; 57 q=q.next; 58 } 59 return w.next; 60 } 61 62 63 64 65 public static class ListNode { 66 int val; 67 ListNode next; 68 ListNode(int x) { val = x; } 69 } 70 }
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
1 package leetcodeTest; 2 import java.util.*; 3 public class addTwoNumbersClass { 4 public static void main(String[]args){ 5 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}; 6 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}; 7 8 ListNode l1=creatList(a); 9 ListNode l2=creatList(b); 10 ListNode l= addTwoNumbers( l1, l2); 11 while(l!=null){ 12 System.out.println(l.val); 13 l=l.next; 14 } 15 } 16 17 public static ListNode addTwoNumbers(ListNode l1, ListNode l2) { 18 if(l1==null&&l2==null) return null; 19 ListNode s = new ListNode(0); 20 ListNode r=s; 21 int m=0; 22 while(l1!=null&&l2!=null){ 23 int a=l1.val; 24 int b=l2.val; 25 int sum=a+b+m; 26 m=sum/10;//向右进位的数 27 int n=sum%10;//当前位 28 ListNode t=new ListNode(n); 29 s.next=t; 30 l1=l1.next; 31 l2=l2.next; 32 s=s.next; 33 } 34 while(l1==null&&l2!=null){ 35 int b=l2.val; 36 int sum=b+m; 37 m=sum/10; 38 int n=sum%10; 39 ListNode t=new ListNode(n); 40 s.next=t; 41 l2=l2.next; 42 s=s.next; 43 } 44 while(l1!=null&&l2==null){ 45 int a=l1.val; 46 int sum=a+m; 47 m=sum/10; 48 int n=sum%10; 49 ListNode t=new ListNode(n); 50 s.next=t; 51 l1=l1.next; 52 s=s.next; 53 } 54 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 55 ListNode t=new ListNode(m); 56 s.next=t; 57 } 58 return r.next; 59 } 60 61 public static ListNode creatList(int[] x){ 62 //创建链表 63 ListNode q = new ListNode(0); 64 ListNode w=q; 65 for(int i=0;i<x.length;i++) { 66 ListNode p = new ListNode(x[i]); 67 q.next = p; 68 q=q.next; 69 } 70 return w.next; 71 } 72 73 74 75 76 public static class ListNode { 77 int val; 78 ListNode next; 79 ListNode(int x) { val = x; } 80 } 81 }
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！
1 class Solution { 2 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 3 ListNode l3 = new ListNode(0); 4 ListNode p = l1, q = l2, current = l3; 5 int carry = 0; 6 while (p != null || q != null) { 7 int x = (p == null) ? 0 : p.val; 8 int y = (q == null) ? 0 : q.val; 9 int sum = x + y + carry; 10 carry = sum/10; 11 current.next = new ListNode(sum%10); 12 current = current.next; 13 if(p!=null) p = p.next; 14 if(q!=null) q = q.next; 15 } 16 if (carry > 0) { 17 current.next = new ListNode(carry); 18 current = current.next; 19 } 20 return l3.next; 21 } 22 }
标签：AC 学习 测试 link 简单 ring bsp sub div
原文地址：https://www.cnblogs.com/pathjh/p/9093592.html
