top100-002-两数相加

此道题出现铁憨憨特征。

题目所给链表,逆序,正符合加法计算从低至高,仅需直接相加,高位不够补0即可。

而自己却多此一举采用队列重新存储一遍,对高位不足时不计入加法中。整体数据结构复杂,并且栽与笨重的固化思维,谨记,以作改变。

 

自己代码:

/**
 * 给出两个 非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。
 * 如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。
 * 您可以假设除了数字 0 之外,这两个数都不会以 0 开头
 * <p>
 * 示例:
 * 输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
 * 输出:7 -> 0 -> 8
 * 原因:342 + 465 = 807
 */

import java.util.Scanner;
import java.util.concurrent.LinkedBlockingQueue;

public class No2 {
    public static void main(String[] args) throws InterruptedException {
        Scanner input = new Scanner(System.in);
        int num1 = input.nextInt();
        int num2 = input.nextInt();
        int temp1 = num1;
        int temp2 = num2;
        ListNode listNode1 = null;
        ListNode nodetemp1 = null;
        ListNode nodetemp2 = null;
        ListNode listNode2 = null;
        while (temp1 > 0) {
            ListNode listNodeTemp = new ListNode(temp1 % 10);
            if (listNode1 == null) {
                listNode1 = listNodeTemp;
                nodetemp1 = listNodeTemp;
            } else {
                nodetemp1.next = listNodeTemp;
                nodetemp1 = listNodeTemp;
            }
            temp1 = temp1 / 10;
        }
        while (temp2 > 0) {
            ListNode listNodeTemp = new ListNode(temp2 % 10);
            if (listNode2 == null) {
                listNode2 = listNodeTemp;
                nodetemp2 = listNodeTemp;
            } else {
                nodetemp2.next = listNodeTemp;
                nodetemp2 = listNodeTemp;
            }
            temp2 = temp2 / 10;
        }
        ListNode newList;
        newList = new No2().addTwoNumbers(listNode1, listNode2);
        while (newList != null) {
            System.out.print(newList.val);
            newList = newList.next;
            if (newList != null) {
                System.out.print(" -> ");
            }
        }
        input.close();
    }

    private ListNode addTwoNumbers(ListNode l1, ListNode l2) throws InterruptedException {
        LinkedBlockingQueue<Integer> queue1 = new LinkedBlockingQueue<>();
        LinkedBlockingQueue<Integer> queue2 = new LinkedBlockingQueue<>();
        while (l1 != null) {
            queue1.put(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            queue2.put(l2.val);
            l2 = l2.next;
        }
        int priority = 0;
        ListNode newList = null;
        ListNode nextNode = null;

        while (!queue1.isEmpty() || !queue2.isEmpty()) {
            int temp = priority;
            while (!queue1.isEmpty()) {
                temp += queue1.poll();
                break;
            }
            while (!queue2.isEmpty()) {
                temp += queue2.poll();
                break;
            }
            if (temp >= 10) {
                priority = temp / 10;
            } else {
                priority = 0;
            }
            ListNode node = new ListNode(temp % 10);
            if (newList != null) {
                nextNode.next = node;
                nextNode = node;
            } else {
                newList = node;
                nextNode = node;
            }
        }
        if (priority > 0) {
            nextNode.next = new ListNode(priority);
        }
        return newList;
    }
}

标准答案:

 

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

 

posted @ 2019-07-26 17:46  Star灬木子李  阅读(98)  评论(0编辑  收藏  举报