题解报告LeetCode#2：两数相加

题意：LeetCode#2：两数相加

思路：

代码：

  1. class Solution {
  2.     public ListNode addTwoNumbers(ListNode l1, ListNodel2) {
  3.         ListNode pre = new ListNode(0);
  4.         ListNode cur = pre;
  5.         int carry = 0;
  6.         int x, y, sum;
  7.         while (l1 != null || l2 != null) {
  8.             x = l1 == null ? 0 : l1.val;
  9.             y = l2 == null ? 0 : l2.val;
  10.             sum  = x + y + carry;
  12.             carry = sum / 10;
  13.             sum = sum % 10;
  14.             cur.next = new ListNode(sum);
  16.             cur = cur.next;
  17.             if (l1 != null)
  18.                 l1 = l1.next;
  19.             if (l2 != null)
  20.                 l2 = l2.next;
  21.         }
  23.         if (carry == 1) {
  24.             cur.next = new ListNode(carry);
  25.         }
  27.         return pre.next;    
  28.     }
  29. }

题解报告LeetCode#445：两数相加Ⅱ

题意：LeetCode#445：两数相加Ⅱ

思路：利用辅助栈

代码：

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> stack1 = new Stack<>();
        Stack<Integer> stack2 = new Stack<>();

        while (l1 != null) {
            stack1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            stack2.push(l2.val);
            l2 = l2.next;
        }

        int carry = 0;
        ListNode head = null;
        while (!stack.isEmpty() || !stack.isEmpty() || carry != 0) {
            int sum = carry;
            sum += stack1.isEmpty() ? 0 : stack1.pop();
            sum += stack2.isEmpty() ? 0 : stack2.pop();
            ListNode node = new ListNode(sum % 10);
            node.next = head;
            head = node;
            carry = sum / 10;
        }

        return head;
    }
}