【LeetCode445】 Add Two Numbers II★★

题目描述：

解题思路：

 　　给定两个链表（代表两个非负数），数字的各位以正序存储，将两个代表数字的链表想加获得一个新的链表（代表两数之和）。

　　如(7->2->4->3)(7243) + (5->6->4)(564) = (7->8->0->7)(7807)

　　此题与【LeetCode2】Add Two Numbers解决思路类似，不同之处在于此题的数字以正序存储，故需要借助栈。

Java代码：

import java.util.Stack;
//public class LeetCode445为测试代码
public class LeetCode445 {
    public static void main(String[] args) {
        ListNode l1=new ListNode(7),l11=new ListNode(2),l12=new ListNode(4),l13=new ListNode(3);
        l1.next=l11;
        l11.next=l12;
        l12.next=l13;
        System.out.print("Input:["+l1.val+","+l11.val+","+l12.val+","+l13.val+"]");
        ListNode l2=new ListNode(5),l21=new ListNode(6),l22=new ListNode(4);
        l2.next=l21;
        l21.next=l22;
        System.out.println(",["+l2.val+","+l21.val+","+l22.val+"]");
        ListNode list=new Solution().addTwoNumbers(l1, l2);
        if(list!=null)
            System.out.print("output：["+list.val);
        while(list.next!=null){
            System.out.print(","+list.next.val);
            list.next=list.next.next;
        }
        System.out.println("]");
    }
}
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> stack1=new Stack<Integer>();
        Stack<Integer> stack2=new Stack<Integer>();
        while(l1!=null){
            stack1.push(l1.val);
            l1=l1.next;
        }
        while(l2!=null){
            stack2.push(l2.val);
            l2=l2.next;
        }
        int carry=0;
        ListNode resultList=new ListNode(0);
        while(!stack1.empty()||!stack2.empty()){
            int sum=carry;
            if(!stack1.empty()) sum+=stack1.pop();
            if(!stack2.empty()) sum+=stack2.pop();
            resultList.val=sum%10;
            ListNode newHead=new ListNode(sum/10);
            newHead.next=resultList;
            resultList=newHead;
            carry=sum/10;
        }
        return resultList.val==0?resultList.next:resultList;
    }
}
class ListNode {
       int val;
       ListNode next;
       ListNode(int x) { val = x; }
      }

程序结果：