top100-002-兩數相加

此道題出現鐵憨憨特徵。

題目所給鏈表，逆序，正符合加法計算從低至高，僅需直接相加，高位不夠補0便可。

而本身卻畫蛇添足採用隊列從新存儲一遍，對高位不足時不計入加法中。總體數據結構複雜，而且栽與笨重的固化思惟，謹記，以做改變。

 

本身代碼：

/**
 * 給出兩個 非空 的鏈表用來表示兩個非負的整數。其中，它們各自的位數是按照 逆序 的方式存儲的，而且它們的每一個節點只能存儲 一位 數字。
 * 若是，咱們將這兩個數相加起來，則會返回一個新的鏈表來表示它們的和。
 * 您能夠假設除了數字 0 以外，這兩個數都不會以 0 開頭
 * <p>
 * 示例：
 * 輸入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
 * 輸出：7 -> 0 -> 8
 * 緣由：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;
    }
}

標準答案：

 

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;
}