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  • 【LeetCode】Add Two Numbers

    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

    按照定义做,逐位相加,注意进位。

    /**
     * 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) {
            ListNode* newhead = new ListNode(-1);
            ListNode* tail = newhead;
            int carry = 0;
            int sum;
            int val;
            while(l1 != NULL && l2 != NULL)
            {
                sum = l1->val + l2->val + carry;
                val = sum%10;
                carry = sum/10;
                tail->next = new ListNode(val);
                tail = tail->next;
                
                l1 = l1->next;
                l2 = l2->next;
            }
            if(l1 == NULL)
            {
                if(l2 == NULL)
                {//l1, l2 == NULL
                    if(carry != 0)
                        tail->next = new ListNode(carry);
                }
                else
                {//l1 == NULL, l2 != NULL
                    while(l2 != NULL)
                    {
                        sum = l2->val + carry;
                        val = sum%10;
                        carry = sum/10;
                        tail->next = new ListNode(val);
                        tail = tail->next;
                        l2 = l2->next;
                    }
                    if(carry != 0)
                        tail->next = new ListNode(carry);
                }
            }
            else
            {//l1 != NULL, that means l2 == NULL
                while(l1 != NULL)
                {
                    sum = l1->val + carry;
                    val = sum%10;
                    carry = sum/10;
                    tail->next = new ListNode(val);
                    tail = tail->next;
                    l1 = l1->next;
                }
                if(carry != 0)
                    tail->next = new ListNode(carry);
            }
            return newhead->next;
        }
    };

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  • 原文地址:https://www.cnblogs.com/ganganloveu/p/4179793.html
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