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  • 2. 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

    ============

    题目:两个链表从head节点对齐,开始按节点相加,并向右进位

    思路:

    设置一个进位标志carray  = 0,两个链表节点指针l1,l2分别指向list1和list2

    当l1和l2的当前位置都有节点时,将两个节点的值相加放入list1中,

    当list1比list2长时,对list1中剩余节点单独判断carray进位标志

    反之,当list2比list1长时,对list2中的剩余节点单独判断carray进位标志

    最后还要判断carray==1?   决定是否new一个新的节点

    code如下:

    /**
     * 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) {
            if(l1==nullptr) return l2;
            if(l2==nullptr) return l1;
            int carray = 0;
            ListNode *prev = nullptr;
            ListNode *head = l1;
            //showList(l1);
            //showList(l2);
            while(l1 && l2){
                int tmp = l1->val + l2->val + carray;
                carray = tmp>9 ? 1:0;
                l1->val = tmp%10;
                prev = l1;
                l1 = l1->next;
                l2 = l2->next;
            }
            while(l1){
                int tmp = l1->val+carray;
                carray = tmp>9 ? 1:0;
                l1->val = tmp%10;
                prev = l1;
                l1 = l1->next;
            }
            while(l2){
                int tmp = l2->val+carray;
                carray = tmp>9 ? 1:0;
                l2->val = tmp%10;
                prev->next = l2;
                prev = l2;
                l2 = l2->next;
            }
            if(carray==1) prev->next = new ListNode(1);
            return head;
        }
    };
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  • 原文地址:https://www.cnblogs.com/li-daphne/p/5607069.html
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