LeetCode445---Add Two Numbers II

topic description

给定两个非空链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储单个数字。将这两数相加会返回一个新的链表。

 

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

进阶:

如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

示例:

输入: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出: 7 -> 8 -> 0 -> 7

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/add-two-numbers-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

  This question is relatively simple, just pay attention to handling the boundary conditions. Because the linked lists are not empty, there is no need for special processing. First, the lengths len1 and len2 of the two chains and the difference dis are obtained, and a stack is used to temporarily store the addition results. Let the longer chain enter the dis position first, and then let the corresponding position values ​​​​of the two chains be added and put in. The previous long chain enters the dis position, which is exactly right at this time. When we finish adding, we get an uncarried addition result. Then slowly pop out the stack, and judge whether there has been a carry here, and the result of the carry needs to be calculated additionally. The result after popping uses the head interpolation method to generate a new chain, and then returns the last node.

ShowCode

/**
 * 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) {
        //先计算两者长度，然后谁长谁先进栈。到了同位的时候就一起进栈。进完后一个个出栈，用头插法生成新的链表
        int len1 = 0,len2 = 0;
        int maxL,dis,minL;
        vector<int> res;
        ListNode*p = l1,*q = l2;
        while(p)
        {
            ++len1;
            p = p->next;
        }
        while(q)
        {
            ++len2;
            q = q->next;
        }
        if(len1 > len2)
        {
            p = l1;
            q = l2;
        }
        else
        {
            p = l2;
            q = l1;
        }
        maxL = len1>len2?len1:len2;
        minL = len1>len2?len2:len1;
        dis = maxL - minL;
        //拿到了两者的长度l1和l2，较长的为maxL,且先移动maxL-min(l1,l2)位
        for(int i = 0;i < dis;++i)
        {
            res.push_back(p->val);
            p = p->next;
        }
        for(int i = 0;i < minL;++i)//相加再进栈
        {
            res.push_back(p->val+q->val);
            p = p->next;
            q = q->next;
        }
        //全部都放进去了，再根据这个vector来还原
        p = NULL;
        while(!res.empty())
        {
            int num = res[res.size()-1];
            res.pop_back();
            cout<<num<<endl;
            if(num > 9)
            {
                if(!res.empty())
                {
                    num -= 10;
                    res[res.size()-1] += 1;   
                }
                else
                {
                    num = num - 10;
                    res.push_back(1);//最后的挣扎;进位
                }
            }
            q = new ListNode(num);
            //新节点是这个，让这个新节点指向前一个节点，并更新前一个节点
            q->next = p;
            p = q;
        }
        return p;
    }
};