lintcode(221)链表求和 II

Description:

假定用一个链表表示两个数，其中每个节点仅包含一个数字。假设这两个数的数字顺序排列，请设计一种方法将两个数相加，并将其结果表现为链表的形式。
Explanation:

给出 6->1->7 +
2->9->5。即，617 + 295。
返回 9->1->2。即，912 。
Solution:

利用栈的先进后出，记录两个链表的节点值。然后计算对应位置的两个数字之和，如果有进位，赋值给flag并带入下一个计算。

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
public ListNode addLists2(ListNode l1, ListNode l2) {
// write your code here
Stack<Integer> temp1 = reverseNode(l1);
Stack<Integer> temp2 = reverseNode(l2);
ListNode point = new ListNode(0);
int flag = 0;
while((!temp1.isEmpty()) && (!temp2.isEmpty())){
int value = temp1.pop() + temp2.pop() + flag;
flag = value/10;
value = value%10;
ListNode cur = new ListNode(value);
cur.next = point.next;
point.next = cur;
}
while(!temp1.isEmpty()){
int value = temp1.pop() + flag;
flag = value/10;
value = value%10;
ListNode cur = new ListNode(value);
cur.next = point.next;
point.next = cur;
}
while(!temp2.isEmpty()){
int value = temp2.pop() + flag;
flag = value/10;
value = value%10;
ListNode cur = new ListNode(value);
cur.next = point.next;
point.next = cur;
}
if(flag == 1){
ListNode cur = new ListNode(1);
cur.next = point.next;
point.next = cur;
}
return point.next;
}

public Stack<Integer> reverseNode(ListNode temp){
Stack<Integer> record = new Stack<Integer>();
while(temp != null){
record.push(temp.val);
temp = temp.next;
}
return record;
}
}