Question: Add Two Numbers

Detail: You are given two non-empty linked lists representing two non-negative integers. 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. You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

• Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
• Output: 7 -> 0 -> 8
• Explanation: 342 + 465 = 807.

 Solution Code:

class Solution {
    /**
     * 需要注意一下幾點：
     * 連結串列長度不相等
     * 進位情況，尤其是進位出現在最後一位
     */
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode
 
* pSumHead = NULL;
        ListNode* pSum = NULL;
        ListNode* p1 = l1;
        ListNode* p2 = l2;
        bool IsHead = true;
        int carry = 0, sum = 0, num = 0;
        // 處理公共長度部分
        while(p1 != NULL && p2 != NULL){
            sum = carry + p1->val + p2->val;
            carry 
 
= (sum >= 10)? 1:0;
            num = sum - 10 * carry;
            ListNode* node = new ListNode(num);
            if(IsHead){// 該節點為head節點
                pSum = node;
                pSumHead = pSum;
                IsHead = false;
            }
            else{
                pSum->next = node;
                pSum = pSum->next;
            }
            p1 = p1->next;
            p2 = p2->next;
        }
        // 兩個連結串列長度相等，且末尾含有進位的情況
        if(p1 == NULL && p2 == NULL && carry == 1)
            pSum->next = new ListNode(carry);
        // 兩個連結串列長度不相等的情況，L2的長度大於L1
        if(p1 == NULL && p2 != NULL){
            while(p2 != NULL){
                sum = carry + p2->val;
                carry = (sum >= 10)? 1:0;
                num = sum - 10 * carry;
                pSum->next = new ListNode(num);
                pSum = pSum->next;
                p2 = p2->next;
            }
            if(carry == 1)
                pSum->next = new ListNode(carry);
        }
        // 兩個連結串列長度不相等的情況，L2的長度小於L1
        if(p1 != NULL && p2 == NULL){
            while(p1 != NULL){
                sum = carry + p1->val;
                carry = (sum >= 10)? 1:0;
                num = sum - 10 * carry;
                pSum->next = new ListNode(num);
                pSum = pSum->next;
                p1 = p1->next;
            }
            if(carry == 1)
                pSum->next = new ListNode(carry);
        }
        return pSumHead;
    }
};

Reports:

Runtime: 28 ms, faster than 98.14% of C++ online submissions for Add Two Numbers.