Add two numbers

最新推荐文章于 2024-08-12 10:14:14 发布

原创最新推荐文章于 2024-08-12 10:14:14 发布 · 407 阅读

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本文深入探讨了如何使用C/C++语言通过单链表进行两个整数相加的操作，包括处理进位并确保结果链表的正确构建。

**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (l1==NULL && l2==NULL)
            return NULL;
        ListNode *retHead = NULL;
        ListNode **res=&retHead;
        int carry = 0;
        int value = 0;
        while (l1!=NULL || l2!= NULL) {
            value=0;
            value +=carry;
            if (l1!=NULL) {
                value += l1->val;
                l1=l1->next;
            }
            if (l2!=NULL) {
                value += l2->val;
                l2=l2->next;
            }
            carry = value/10;
            value = value;
               
            *res = new ListNode(value);
            res =  &(*res)->next;
        }
        if (carry >0)
            *res = new ListNode(carry);
        return retHead;
    }
};

14/1/10: 注意最后一位的进位，可能比原链表长

/**
 * 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==NULL && l2==NULL) return NULL;
        ListNode *pre = new ListNode(0);
        ListNode *res=pre;
        int c=0;
        int s=0;
        while (l1 || l2) {
            if (l1 && l2) {
                s=l1->val +l2->val+c;
                l1=l1->next;
                l2=l2->next;
            }else if (!l2) {
                s=l1->val+c;
                l1=l1->next;
            }else {
                s=+l2->val+c;
                l2=l2->next;
            }
            c=s/10;
            s=s%10;
            ListNode *cur = new ListNode(s);
            pre->next = cur;
            pre=cur;
        }
        if (c>0) {
            ListNode *cur = new ListNode(c);
            pre->next = cur;
        }
        return res->next;
    }
};