leetcode: Add Two Numbers

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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

思路非常简单，利用两个指针分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
//        ListNode *pResult = NULL;
//        ListNode **pCur = &pResult;

        ListNode rootNode(0);
        ListNode *pCurNode = &rootNode;
        int a = 0;
        while (l1 || l2)
        {
            int v1 = (l1 ? l1->val : 0);
            int v2 = (l2 ? l2->val : 0);
            int temp = v1 + v2 + a;
            a = temp / 10;
            ListNode *pNode = new ListNode((temp % 10));
            pCurNode->next = pNode;
            pCurNode = pNode;
            if (l1)
                l1 = l1->next;
            if (l2)
                l2 = l2->next;
        }
        if (a > 0)
        {
            ListNode *pNode = new ListNode(a);
            pCurNode->next = pNode;
        }
        return rootNode.next;
    }
};

leetcode: Add Two Numbers

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原文地址：http://blog.csdn.net/flyljg/article/details/47658687