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
思路:模擬加法。用兩個指針分別指向兩個鏈表,進行相加,用一個變量記錄進位情況。如果有鏈表為空,則處理另一個,直到兩個鏈表都處理完為止。
實現代碼:
/** * 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 preHead(0); ListNode *p = &preHead; int carry = 0; while (l1 || l2 || carry) { int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry; carry = sum / 10; p->next = new ListNode(sum % 10); p = p->next; l1 = l1 ? l1->next : l1; l2 = l2 ? l2->next : l2; } return preHead.next; } };