两数相加--链表加法--LeetCode练习

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题目描述

给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0 开头。

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

题解

• 其实就是考验对于cpp指针的理解而已啦。
• 注意下细节就好了

/**
 * 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) {
        int tmp = 0, flag = 0;
        ListNode * head = NULL, ** curr = NULL;
        while(l1 != NULL && l2 != NULL){
            tmp = l1->val + l2->val + flag;
            if (head == NULL) { 
                head = new ListNode(tmp % 10);
                curr = &head;
            } else {
                (*curr) = new ListNode(tmp % 10);
            }
            flag = tmp / 10;
            
            curr = &((*curr)->next);
            l1 = l1 -> next;
            l2 = l2 -> next;
        }
        while(l1 != NULL) {
            tmp = l1->val + flag;
            *curr = new ListNode(tmp % 10);
            flag = tmp / 10;
            if (head == NULL) { 
                head = new ListNode(tmp % 10);
                curr = &head;
            } else {
                (*curr) = new ListNode(tmp % 10);
            }
            curr = &((*curr)->next);
            l1 = l1->next;
        }
        
        while(l2 != NULL) {
            tmp = l2->val + flag;
            *curr = new ListNode(tmp % 10);
            flag = tmp / 10;
            if (head == NULL) { 
                head = new ListNode(tmp % 10);
                curr = &head;
            } else {
                (*curr) = new ListNode(tmp % 10);
            }
            curr = &((*curr)->next);
            l2 = l2->next;
        }
        while (flag != 0) {
            tmp = flag;
            *curr = new ListNode(tmp % 10);
            flag = tmp / 10;
            if (head == NULL) { 
                head = new ListNode(tmp % 10);
                curr = &head;
            } else {
                (*curr) = new ListNode(tmp % 10);
            }
            curr = &((*curr)->next);
        }
        return head;
    }
};

看到了大佬的一份用java写的程序，看起来冗余更少，更精致。我改写成cpp学习下

/**
 * 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) {
        int tmp = 0, flag = 0, x, y;
        ListNode * head = NULL, ** curr = NULL;
        while(l1 != NULL || l2 != NULL || flag){
            x = (l1 == NULL) ? 0: l1 -> val;
            y = (l2 == NULL) ? 0: l2 -> val;
            tmp = x + y + flag;
            if (head == NULL) { 
                head = new ListNode(tmp % 10);
                curr = &head;
            } else {
                (*curr) = new ListNode(tmp % 10);
            }
            flag = tmp / 10;
            
            curr = &((*curr)->next);
            l1 = (l1 == NULL) ? NULL: l1 -> next;
            l2 = (l2 == NULL) ? NULL: l2 -> next;
        }
        return head;
    }
};