leetcode-445-Add-Two-Numbers-II

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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:

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Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

分析

两个链表相加，然而最高位在链表首位，题目又要求不能翻转链表，可以用先进后出队列解决。

解决方案1(Java)

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carry = 0;
        Deque<ListNode> deque1 = new LinkedList<>(), deque2 = new LinkedList<>();
        ListNode current1 = l1, current2 = l2;
        while (current1 != null || current2 != null) {
            if (current1 != null) {
                deque1.push(current1);
                current1 = current1.next;
            }
            if (current2 != null) {
                deque2.push(current2);
                current2 = current2.next;
            }
        }
        
        ListNode head = new ListNode(0), tmp = null;
        while(!deque1.isEmpty() || !deque2.isEmpty()) {
            int n1 = deque1.isEmpty()? 0: deque1.pop().val;
            int n2 = deque2.isEmpty()? 0: deque2.pop().val;
            int sum = n1 + n2 + carry;
            carry = sum / 10;
            head.val = sum % 10;
            tmp = new ListNode(0);
            tmp.next = head;
            head = tmp;
        }
        if (carry == 1) {
            head.val = 1;
        }
        return carry == 1 ? head: head.next;
    }
}

相关问题

题目来源