链表求和

leetCode:

https://leetcode.com/problems/add-two-numbers-ii/description/

 

描述：

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


这题最容易想到的思路就是将链表反转，然后同时遍历两个链表，最后再将加和后的链表反转回来。
但是题目最后提高了难度，如果原链表不能改变，也就是不能反转链表，该如何解决呢？？

答案就是使用栈，遍历链表依次将元素压入栈中，元素的顺序就被反转过来了，低位的数字在栈上面，高位的数字在栈底部。
循环第从两个栈顶弹出元素，执行加法逻辑，并将相加后的元素放入一个新的栈中，这样求和结果中低位数字在栈底，高位数字在栈顶，

最后循环从栈顶弹出元素，构建节点之间的指向关系，形成从高位到低位的链表。

代码：

public class AddTwoNumbers {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> stack1 = pushToStck(l1);
        Stack<Integer> stack2 = pushToStck(l2);
        Stack<ListNode> result = new Stack<>();
        int carry = 0;
        while (true) {
            if (stack1.isEmpty() && stack2.isEmpty()) {
                break;
            }
            int s1 = stack1.isEmpty() ? 0 : stack1.pop();
            int s2 = stack2.isEmpty() ? 0 : stack2.pop();
            result.push(new ListNode((s1 + s2 + carry) % 10));
            carry = (s1 + s2 + carry) / 10;
        }
        if (carry != 0) {
            result.push(new ListNode(1));
        }
        ListNode newHead = null, newTail = null;
        while (!result.isEmpty()) {
            if (newHead == null) {
                newHead = newTail = result.pop();
            } else {
                newTail.next = result.pop();
                newTail = newTail.next;
            }
        }
        return newHead;
    }

    public Stack<Integer> pushToStck(ListNode l) {
        Stack<Integer> stack = new Stack<>();
        while (l != null) {
            stack.push(Integer.valueOf(l.val));
            l=l.next;
        }
        return stack;
    }
}