标签：
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
题意：将两个用链表逆序存储的数进行加法运算，并将结果也用链表逆序存储。如上例的342 + 564 = 807.
题解：掌握链表的头插法和尾插法，从左往右每个节点进行加法运行，进位带入下个个节点运算。用尾插法保存结果数据。要注意节点的空判断，以防出现空指针，当计算最后的进位不为0时，要再加入一个节点存放进位。
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
public class AddTwoNumbers {
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode res = new ListNode(0);
ListNode r = res;
int x = l1.val;
int y = l2.val;
res.val = ( x + y ) % 10;
int ten = ( x + y ) / 10;
l1 = l1.next;
l2 = l2.next;
while(l1 != null || l2 !=null ){
// 防止出现空指针
if(l1 == null){
l1 = new ListNode(0);
x = 0;
}else{
x = l1.val;
}
if(l2 == null){
l2 = new ListNode(0);
y = 0;
}else{
y = l2.val;
}
int single = ( x + y + ten ) % 10;
ListNode tmp = new ListNode(single);
r.next = tmp;
r = tmp;
ten = ( x + y + ten ) / 10;
l1 = l1.next;
l2 = l2.next;
}
//存放最后的进位
if(ten != 0){
r.next = new ListNode(ten);
}
return res;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int a[] = {5};
int b[] = {5};
ListNode la = new ListNode(0);
la.val = a[0];
for(int i = a.length -1 ;i>0;i--){
ListNode next = new ListNode(a[i]);
next.next = la.next;
la.next = next;
}
ListNode lb = new ListNode(0);
lb.val = b[0];
for(int i = b.length -1 ;i > 0 ;i--){
ListNode next = new ListNode(b[i]);
next.next = lb.next;
lb.next = next;
}
ListNode c = addTwoNumbers(la,lb);
while( c != null){
System.out.print(c.val);
if(c.next != null){
System.out.print("-->");
}
c = c.next;
}
}
}
标签：
原文地址：http://www.cnblogs.com/AndyDai/p/4810864.html
