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[LeetCode] Add Two Numbers
You 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
/** * 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 = l1; int flag = 0; ListNode *li1 = new ListNode(0); ListNode *li2 = new ListNode(0); li1->next = l1; li2->next = l2; while(li1->next!=NULL && li2->next!=NULL) { li1->next->val = li1->next->val+li2->next->val+flag; if(li1->next->val>9) { li1->next->val = li1->next->val-10; flag = 1; } else { flag = 0; } li1 = li1->next; li2 = li2->next; } while(li1->next != NULL && li2->next == NULL) { if(flag==1 && li1->next->val<9) { ++li1->next->val; flag = 0; } else if(flag==1 && li1->next->val==9) { li1->next->val = 0; flag = 1; } li1=li1->next; } while(li2->next != NULL && li1->next == NULL) { ListNode *p1; if(flag==1 && li2->next->val<9) { p1 = new ListNode(li2->next->val+1); flag = 0; } else if(flag==1 && li2->next->val==9) { p1 = new ListNode(0); flag = 1; } else { p1 = new ListNode(li2->next->val); } li1->next =p1; li1 = li1->next; li2 = li2->next; } if(flag == 1) { ListNode *p = new ListNode(1); li1->next = p; } return head; } };
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