• 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

    思路:
    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。
    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
            int carry = 0;  
            ListNode* tail = new ListNode(0);  
            ListNode* ptr = tail;  
              
            while(l1 != NULL || l2 != NULL){  
                int val1 = 0;  
                if(l1 != NULL){  
                    val1 = l1->val;  
                    l1 = l1->next;  
                }  
                  
                int val2 = 0;  
                if(l2 != NULL){  
                    val2 = l2->val;  
                    l2 = l2->next;  
                }  
                  
                int tmp = val1 + val2 + carry;  
                ptr->next = new ListNode(tmp % 10);  
                carry = tmp / 10;  
                ptr = ptr->next;  
            }  
              
            if(carry == 1){  
                ptr->next = new ListNode(1);  
            }  
            return tail->next;  
            
            
        }
    };
     
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  • 原文地址:https://www.cnblogs.com/qiaozhoulin/p/4574704.html
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