# 9. 两数相加

java

```java
//------Link工具类------
public class LinkedUtils {

    /**
     * 数组转化为链表
     * @param array
     * @return
     */
    public static ListNode arrayToLinkedList(int[] array){
        if(array==null) return null;
        //head node 
        ListNode root=new ListNode(-1);
        ListNode p=root;

        int size=array.length;
        int i=0;
        while(i<size){
            ListNode q=new ListNode(array[i++]);
            q.next=null;
            p.next=q;
            p=q;
        }

        return root.next;
    }

    /**
     * 打印链表
     * @param root
     */
    public static void print(ListNode root) {
        if(root==null) System.out.println("root is null");
        else{
            while(root!=null){
                System.out.print(root.val);
                if(root.next!=null){
                    System.out.print("->");
                }
                root=root.next;
            }
            System.out.println();
        }
    }

    /**
     * 求链表长度
     * @param root
     * @return
     */
    public static int length(ListNode root) {
        if(root==null) return 0;
        int n=0;
        while(root!=null){
            root=root.next;
            n++;
        }
        return n;
    }

    /**
     * 返回指向链表尾部的指针
     * @param root
     * @return
     */
    public static ListNode moveToTail(ListNode root) {
        if(root==null) return null;
        ListNode tail=root;
        while(tail.next!=null){
            tail=tail.next;
        }
        return tail;
    }
}


//------节点定义类------
public class ListNode {
    public int val;
    public ListNode next;

    public ListNode(int val) {
        this.val=val;
    }
}



//------真正开始问题------

/**
 * 两数相加.
 * https://leetcode-cn.com/problems/add-two-numbers/description/
 *
 * 给定两个非空链表来表示两个非负整数。位数按照逆序方式存储，它们的每个节点只存储单个数字。将两数相加返回一个新的链表。
 * 你可以假设除了数字 0 之外，这两个数字都不会以零开头。
 *
 * 示例：
 * 输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
 * 输出：7 -> 0 -> 8
 * 原因：342 + 465 = 807
 */
public class Solution2 {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode node = new ListNode(-1);//头节点
        ListNode p = node;//指示执行到的位置
        int plusOne = 0;//是否进位

        while (l1 != null || l2 != null) {
            //计算对应位和
            int v1 = (l1 != null) ? l1.val : 0;
            int v2 = (l2 != null) ? l2.val : 0;
            int r = v1 + v2 + plusOne;

            //之和大于9，减去十即为该值，并进位
            if (r > 9) {
                plusOne = 1;
                ListNode tmp = new ListNode(r - 10);
                p.next = tmp;
                p = tmp;
            } else {
                plusOne = 0;
                ListNode tmp = new ListNode(r);
                p.next = tmp;
                p = tmp;
            }
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }

        //如果最后一位仍然需要进位，创建节点
        if (plusOne == 1) {
            p.next = new ListNode(1);
        }
        return node.next;
    }

    public static void main(String[] args) {
        //你的算法必须是高效的，否则下面这个测试集是过不去的
        int[] list1 = {
                2,4,3,2,4,3,2,4,3,2,
                4,3,2,4,3,2,4,3,2,4,
                3,2,4,3,2,4,3,2,4,3,
                2,4,3,2,4,3,2,4,3,2,
                4,3,2,4,3,2,4,3,2,4,
                3,2,4,3,2,4,3,2,4,3,9 };
        int[] list2 = {
                5,6,4,2,4,3,2,4,3,2,
                4,3,2,4,3,2,4,3,2,4,
                3,2,4,3,2,4,3,2,4,3,
                2,4,3,2,4,3,2,4,3,2,
                4,3,2,4,3,2,4,3,2,4,
                3,2,4,3,2,4,3,9,9,9,9 };

        ListNode l1 = LinkedUtils.arrayToLinkedList(list1);
        ListNode l2 = LinkedUtils.arrayToLinkedList(list2);

        ListNode root = new Solution2().addTwoNumbers(l1, l2);
        LinkedUtils.print(root);
    }
}
```

python

```python
```


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