[leetcode] Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

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

https://oj.leetcode.com/problems/add-two-numbers/

思路：模拟题，从头向后依次遍历两个链表相加。

注意：1. 用dummy head简便处理。2. 注意carry的处理，尤其最高位进位的情况。3. 注意引用null的判断。

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        // printList(l1);
        // printList(l2);
        ListNode p = new ListNode(-1);
        ListNode head = p;
        int c = 0;
        while (l1 != null || l2 != null || c == 1) {
            int curInt = 0;
            if (l1 != null)
                curInt += l1.val;
            if (l2 != null)
                curInt += l2.val;
            curInt += c;
            if (curInt > 9) {
                c = 1;
                curInt -= 10;
            } else
                c = 0;
            p.next = new ListNode(curInt);

            p = p.next;
            if (l1 != null)
                l1 = l1.next;
            if (l2 != null)
                l2 = l2.next;
        }

        // printList(head.next);

        return head.next;
    }

    void printList(ListNode node) {
        while (node != null) {
            System.out.print(node.val);
            if (node.next != null)
                System.out.print("->");
            node = node.next;
        }
        System.out.println();

    }

    public static void main(String[] args) {
        ListNode l1 = new ListNode(2);
        l1.next = new ListNode(4);
        l1.next.next = new ListNode(5);

        ListNode l2 = new ListNode(5);
        l2.next = new ListNode(6);
        l2.next.next = new ListNode(4);

        new Solution().addTwoNumbers(l1, l2);

    }

}

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
        next = null;
    }
}

第二遍记录：l1,l2指针忘记往后移动了

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(-1);
        ListNode cur = head;
        int c=0;
        while(l1!=null||l2!=null||c!=0){
            int curVal=0;
            if(l1!=null)
                curVal+=l1.val;
            if(l2!=null)
                curVal+=l2.val;
            curVal+=c;
            
            if(curVal>9){
                curVal-=10;
                c=1;
            }else{
                c=0;
            }
            cur.next= new ListNode(curVal);
            
            cur=cur.next;
            if(l1!=null)
                l1=l1.next;
            if(l2!=null)
                l2=l2.next;
        }
        return head.next;
    }
}

复习注意：

1. dummyhead的使用。

2. carry的处理。

3. 三个指针不要忘记移动。