You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
错误解法：由于数太长。用int，long存不下，导致结果错误。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if(l1==null) return l2;
if(l2==null) return l1;
ListNode node1=l1;
ListNode node2=l2;
int a =0,b=0;
int c=1;
while(node1!=null) {
a+=(c*node1.val);
c=c*10;
node1=node1.next;
}
c=1;
while(node2!=null){
b+=(c*node2.val);
c=c*10;
node2=node2.next;
}
int d=a+b;
String res=Integer.toString(d);
ListNode l3=new ListNode(0);
ListNode node3=l3;
for(int i=res.length()-1;i>=0;i--){
node3.val=(int)(res.charAt(i))-(int)('0');
if(i>0){
node3.next=new ListNode(0);
node3=node3.next;
}
}
return l3;
}
}正确解法。空间效率极差。肯定还有更好的解法，没工夫研究，待续……
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if(l1==null) return l2;
if(l2==null) return l1;
ListNode node1=l1;
ListNode node2=l2;
List<Integer> list1=new ArrayList<Integer>();
List<Integer> list2=new ArrayList<Integer>();
while(node1!=null) {
list1.add(node1.val);
node1=node1.next;
}
while(node2!=null){
list2.add(node2.val);
node2=node2.next;
}
while(list1.size()<list2.size()) list1.add(0);
while(list2.size()<list1.size()) list2.add(0);
ListNode l3=new ListNode(0);
ListNode node3=l3;
int tmp=0;
for(int i=0;i<list1.size();i++){
int a=list1.get(i)+list2.get(i)+tmp;
if(a>=10){
node3.val=a-10;
tmp=1;
}
else{
node3.val=a;
tmp=0;
}
if(i<list1.size()-1){
node3.next=new ListNode(0);
node3=node3.next;
}
}
if(tmp==1){
node3.next=new ListNode(tmp);
}
return l3;
}
}版权声明：本文内容由互联网用户自发贡献，该文观点仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请联系我们举报，一经查实，本站将立刻删除。
发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/109425.html原文链接：https://javaforall.net
