LeetCode 445. 两数相加 II(Add Two Numbers II)

445. 两数相加 II
445. Add Two Numbers II

题目描述
给定两个非空链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储单个数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外,这两个数字都不会以零开头。

进阶:
如果输入链表不能修改该如何处理?换句话说,你不能对列表中的节点进行翻转。

LeetCode445. Add Two Numbers II中等

示例:

输入: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) 输出: 7 -> 8 -> 0 -> 7

Java 实现

public class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
    }

    @Override
    public String toString() {
        return val + "->" + next;
    }
}
import java.util.Stack;

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> s1 = new Stack<Integer>();
        Stack<Integer> s2 = new Stack<Integer>();

        while (l1 != null) {
            s1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            s2.push(l2.val);
            l2 = l2.next;
        }

        int sum = 0;
        ListNode list = new ListNode(0);
        while (!s1.empty() || !s2.empty()) {
            if (!s1.empty()) sum += s1.pop();
            if (!s2.empty()) sum += s2.pop();
            list.val = sum % 10;
            ListNode head = new ListNode(sum / 10);
            head.next = list;
            list = head;
            sum /= 10;
        }
        return list.val == 0 ? list.next : list;
    }
}

测试代码

import java.util.Stack;

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> s1 = new Stack<Integer>();
        Stack<Integer> s2 = new Stack<Integer>();

        while (l1 != null) {
            s1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            s2.push(l2.val);
            l2 = l2.next;
        }

        int sum = 0;
        ListNode list = new ListNode(0);
        while (!s1.empty() || !s2.empty()) {
            if (!s1.empty()) sum += s1.pop();
            if (!s2.empty()) sum += s2.pop();
            list.val = sum % 10;
            ListNode head = new ListNode(sum / 10);
            head.next = list;
            list = head;
            sum /= 10;
            System.out.println(list);
        }
        return list.val == 0 ? list.next : list;
    }

    public static void main(String[] args) {
        ListNode t1 = new ListNode(7);
        ListNode t2 = new ListNode(2);
        ListNode t3 = new ListNode(4);
        ListNode t4 = new ListNode(3);
        t1.next = t2;
        t2.next = t3;
        t3.next = t4;

        ListNode p1 = new ListNode(5);
        ListNode p2 = new ListNode(6);
        ListNode p3 = new ListNode(4);
        p1.next = p2;
        p2.next = p3;

        System.out.println(new Solution().addTwoNumbers(t1, p1));
    }
}

测试结果

0->7->null 1->0->7->null 0->8->0->7->null 0->7->8->0->7->null 7->8->0->7->null

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参考资料

原文地址:https://www.cnblogs.com/hgnulb/p/10963035.html