【JAVA、C++】LeetCode 002 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

解题思路：

定义三个ListNode l1、l2，result，其中result为return语句的输出，l1、l2为传入的参数。

将l1赋值给result，执行result.val+=l2.val，然后l1作为指针一级一级往下走，直到走到l2.next为null。当然，之间会有不少边界条件，自己debug一下就好。

Java代码如下：

public class Solution {
      static public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    	ListNode result=l1;
    	while(true){
    		l1.val+=l2.val;
    		if(l1.val>=10){
    			l1.val%=10;
    			if(l1.next==null) l1.next=new ListNode(1);
    			else l1.next.val+=1;
    		}
    		
		if(l2.next==null){
			ListNode l3=l1.next;
			while(true){
				if (l3==null) break;
				if(l3.val==10){
					l3.val%=10;
	    			if(l3.next==null) l3.next=new ListNode(1);
	    			else l3.next.val+=1;
				}
				l3=l3.next;
			}
			break;
		}
		
    	l2=l2.next;
    	if(l1.next==null){
    		l1.next=new ListNode(0);
    	}
    	l1=l1.next;
    	}
    	return result;
    	
    }
}

 C++

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* res=l1;
        while (true) {
            l1->val += l2->val;
            if (l1->val >= 10) {
                l1->val %= 10;
                if (l1->next == NULL) 
                    l1->next = new ListNode(1);
                else l1->next->val += 1;
            }

            if (l2->next == NULL) {
                ListNode* l3 = l1->next;
                while (true) {
                    if (l3 == NULL) break;
                    if (l3->val == 10) {
                        l3->val %= 10;
                        if (l3->next == NULL) l3->next = new ListNode(1);
                        else l3->next->val += 1;
                    }
                    l3 = l3->next;
                }
                break;
            }

            l2 = l2->next;
            if (l1->next == NULL) {
                l1->next = new ListNode(0);
            }
            l1 = l1->next;
        }
        return res;
    }
};