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
這道題可以看作兩個鏈表的相加，也可以看成342+465=807，只不過換作鏈表的形式。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry =0;
ListNode newHead = new ListNode(0);
ListNode p1 = l1, p2 = l2, p3=newHead;
while(p1 != null || p2 != null){//兩鏈表數(shù)據(jù)相應(yīng)的節(jié)點相加在塔，carry用于儲存之前進(jìn)位的數(shù)據(jù)
if(p1 != null){
carry += p1.val;
p1 = p1.next;
}
if(p2 != null){
carry += p2.val;
p2 = p2.next;
}
p3.next = new ListNode(carry%10);//將運算結(jié)果儲存在newHead的next節(jié)點處
p3 = p3.next;
carry /= 10;//進(jìn)位的值
}
if(carry==1)
p3.next=new ListNode(1);//最后判斷最高位是否進(jìn)一位，如果大于9，就再創(chuàng)建一個節(jié)點。
return newHead.next;
}
}
