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问题
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.
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Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8 -
解决
package com.lsy.leetcode;
public class AddTwoNumbers {
static AddTwoNumbers a=new AddTwoNumbers();
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
// if (l1 == null) { //说了非空
// return l2;
// }
//
// if (l2 == null) {
// return l1;
// }
int carry = 0;
ListNode newhead = a.new ListNode(-1);
ListNode l3 = newhead;
//ListNode l3;
while(l1!=null || l2!=null){
if(l1!=null){
carry += l1.val;
// System.out.println(l1.val+"test");
l1 = l1.next;
}
if(l2!=null){
carry += l2.val;
l2 = l2.next;
}
//l3.next = a.new ListNode(carry%10); //测试不能从l3开始也可以的
l3.val=(carry%10);//那就不是指向newhead了,需要加一个改变val的方法
System.out.println(newhead.val);
carry = carry/10;
l3.next=a.new ListNode(-1);//用之前实例化l3;
l3=l3.next; //不能和上一行调换位置
}
if(carry == 1)
l3.next=a.new ListNode(1);
// return newhead.next;
// System.out.println(newhead.val+"->"+newhead.next.val+"->"+newhead.next.next.val);
return newhead;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
ListNode l1=a.new ListNode(2); //内部类,不懂呢
l1.next=a.new ListNode(4);
l1.next.next=a.new ListNode(3);
ListNode l2=a.new ListNode(5); //内部类,不懂呢
l2.next=a.new ListNode(6);
l2.next.next=a.new ListNode(4);
//l1.next=new ListNode(4);
ListNode l3=addTwoNumbers(l1,l2);
System.out.println(l3.val+"->"+l3.next.val+"->"+l3.next.next.val);
System.out.println(l1.val); //测试引用传递,l1没有变。传递的是引用的复制。
//参考:http://www.cnblogs.com/lixiaolun/p/4311863.html
}
class ListNode {
int val;
ListNode next;
public ListNode(int x) { val = x; }
}
public ListNode addTwoNumbers2(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; //先确定下来
int y = (q != null) ? q.val : 0;
int sum = carry + x + y;
carry = sum / 10; //取整
curr.next = new ListNode(sum % 10); //取余
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;
}
}
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