2. Add Two Numbers
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Add Two Numbers
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
exaple:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8
思路:
理解题目为,数字两两相加,并朝下一位进位。
情况一:其中一个列表提前加完,将没加完的补充到需要输出的列表处。
情况二:同时加完,却仍有进位,需创建一个结点存储。
代码:
/** * 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 up = 0; ListNode *pNode = NULL, *pNext = NULL, *resList = NULL; //定义ListNode类型的指针 ListNode *p = l1, *q = l2; while (p != NULL && q != NULL){ pNext = new ListNode(q->val + p->val + up);//将指针指向新的结点 up = pNext->val / 10; pNext->val = pNext->val % 10; if(resList == NULL){ resList = pNode = pNext;//保持头结点的不动 } else{ pNode->next = pNext;//结点连接 pNode = pNext; } p = p->next; q = q->next; } while (q != NULL){ //l2仍有剩余 pNext = new ListNode(q->val + up); up = pNext->val / 10; pNext->val = pNext->val % 10; pNode->next = pNext;//结点连接 pNode = pNext; q = q->next; } while (p != NULL){ //l1仍有剩余 pNext = new ListNode(p->val + up); up = pNext->val / 10; pNext->val = pNext->val % 10; pNode->next = pNext;//结点连接 pNode = pNext; p = p->next; } if (up != 0){ //都没有剩余,却仍有进位,需要再创建一个点 pNext = new ListNode(up); pNode->next = pNext; pNode = pNext; } return resList; //返回结果的头指针 }};
知识点:指针、列表的使用,指针的向下移动
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