Add Two Numbers

最新推荐文章于 2024-05-07 14:18:04 发布

原创最新推荐文章于 2024-05-07 14:18:04 发布 · 505 阅读

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本文介绍了一个解决两数相加问题的链表实现方案。通过遍历两个链表，将对应节点的值相加并考虑进位，形成新的链表节点。文章详细展示了如何处理不同长度链表的加法运算，并在最后正确处理可能存在的进位。

简单题

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
         ListNode *head = NULL;
        ListNode *preNode = NULL;
        int value = 0, carrier = 0;
        while(l1 != NULL && l2 !=NULL) {
            value = (l1->val + l2->val + carrier) % 10;
            carrier = (l1->val + l2->val + carrier) / 10;
            ListNode *curNode = new ListNode(value);
            if(head == NULL) {
                head = curNode;
            }
            if(preNode != NULL) {
                preNode->next = curNode;
            }
            preNode = curNode;
            l1 = l1->next;
            l2 = l2->next;
        }

        while(l1 != NULL) {
            value = (l1->val + carrier) % 10;
            carrier = (l1->val + carrier) / 10;
            ListNode *curNode = new ListNode(value);
            if(head == NULL) {
                head = curNode;
            }
            if(preNode != NULL) {
                preNode->next = curNode;
            }
            preNode = curNode;
            l1 = l1->next;
        }
        
        while(l2 != NULL) {
            value = (l2->val + carrier) % 10;
            carrier = (l2->val + carrier) / 10;
            ListNode *curNode = new ListNode(value);
            if(head == NULL) {
                head = curNode;
            }
            if(preNode != NULL) {
                preNode->next = curNode;
            }
            preNode = curNode;
            l2 = l2->next;
        }
        if(carrier > 0) {
            ListNode *curNode = new ListNode(carrier);
            preNode->next = curNode;
        }
        return head;
    }
};