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【Leet Code】Add Two Numbers
Add Two Numbers
Total Accepted: 20255 Total Submissions: 88115My SubmissionsYou are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
这道题目主要考察对创建链表的运用而已,注意:表头是个位:
/** * 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) { ListNode* head = NULL; ListNode* cur = NULL; int ad = 0; while(l1 && l2) { if(NULL ==head) { head = cur = new ListNode( (l1->val + l2->val + ad) % 10 ); } else { cur->next = new ListNode( (l1->val + l2->val + ad) % 10 ); cur = cur->next; } ad = (l1->val + l2->val + ad ) / 10; l1 = l1->next; l2 = l2->next; } while(l1) { cur->next = new ListNode( (l1->val + ad) % 10 ); ad = (l1->val + ad) / 10; cur = cur->next; l1 = l1->next; } while(l2) { cur->next = new ListNode( (l2->val + ad) % 10 ); ad = (l2->val + ad) / 10; cur = cur->next; l2 = l2->next; } if (ad > 0) { cur->next = new ListNode(ad); } return head; } };再附上大神的代码:
class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { // Start typing your C/C++ solution below // DO NOT write int main() function ListNode *ret = NULL; ListNode **pCur = &ret; int nxt = 0; while (l1 && l2) { *pCur = new ListNode((l1->val + l2->val + nxt) % 10); nxt = (l1->val + l2->val + nxt) / 10; pCur = &((*pCur)->next); l1 = l1->next; l2 = l2->next; } while (l1 != NULL) { *pCur = new ListNode((l1->val + nxt) % 10); nxt = (l1->val + nxt) / 10; pCur = &((*pCur)->next); l1 = l1->next; } while (l2 != NULL) { *pCur = new ListNode((l2->val + nxt) % 10); nxt = (l2->val + nxt) / 10; pCur = &((*pCur)->next); l2 = l2->next; } if (nxt > 0) { *pCur = new ListNode(nxt); } return ret; } };
【Leet Code】Add Two Numbers
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