LeetCode-Add Two Numbers II

Description

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 8 -> 0 -> 7

Solution

思路：

这里使用了两个栈将两个链表的值反置；然后一次按位相加即可得到和。

这里有几个点需要注意下：

1. 相加后得到的数据是从低位开始的。需要进行反置(或者直接在返回值上做文章)

2. 按位相加得到和后，需要注意下最后是否还需要高位进一；如果需要则需要补充一个高位。

/**
* 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) {
Stack<Integer> sta1 = new Stack<Integer>();
Stack<Integer> sta2 = new Stack<Integer>();
while (l1 != null) {
sta1.push(l1.val);
l1 = l1.next;
}
while (l2 != null) {
sta2.push(l2.val);
l2 = l2.next;
}

ListNode re  = null;
int x = 0;
while (!sta1.isEmpty() || !sta2.isEmpty()) {
x += (sta1.isEmpty() ? 0 : sta1.pop()) + (sta2.isEmpty() ? 0 : sta2.pop());
ListNode node = new ListNode(x%10);
node.next = re;
re = node;
x = x / 10;
}
if(x ==1){
ListNode node = new ListNode(1);
node.next = re;
re = node;
}
return re;
}
}