题解 | #链表相加(二)#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:

    int getListNodesize(ListNode *node) {
        int count = 0;
        while (node != nullptr) {
           count++;
           node=node->next;
        }
        return count;
    }

    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    ListNode* addInList(ListNode* head1, ListNode* head2) {
        // write code here
        ListNode *l = head1 , *r = head2;
        int ls = getListNodesize(l);
        int rs = getListNodesize(r);
        int lv[1000000]= {0};
        int rv[1000000]= {0};
        int size = ls;
        if (ls < rs) {
            ListNode *tmp = head1;
            head1 = head2;
            head2 = tmp;
            
            size = rs;
        }
        int diff = abs(ls - rs);
        for (int i=1; i<= size; ++i) {
            lv[i] = head1->val;
            head1 = head1->next;
        }
        for (int i=diff + 1; i<=size; ++i) {
            rv[i] = head2->val;
            head2 = head2->next;
        }
        // add 
        int ts = size;
        while (size > 0) {
            int value = lv[size] + rv[size];
            if (value > 9) {
                lv[size - 1]++;
            }
            lv[size] = value % 10;
            size--;
        }

        ListNode* head = new ListNode(-1);
        head->next =nullptr;
        ListNode *cur = head;
        for (int i=0; i<= ts; i++) {
            if (i==0 && lv[i] == 0) {
                continue;
            }
            ListNode* tmp = new ListNode(lv[i]);
            tmp->next = cur->next;
            cur->next =tmp;
            cur = cur->next;
        }

        return head->next;
    }
};

简单记录一下不一样的思路，通过大数加法的办法，处理好链表数据到数组；进行加法后在生成所需的链表；减少了链表的操作，但是增加了空间的消耗；