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
这题虽然归类到Medium 但是难度不大,只是要考虑的东西有点多,比如:两个list不一定一样长,进位这两个问题。
因为两个数字是逆序存到链表里的,所以这大大简化了问题难度,因为只需要向着链表遍历方向往前进位就可以了。
void addIter(ListNode *l1, ListNode *l2, ListNode *prev, int carry) { if (!l1 && !l2 && carry == 0) return; ListNode *node; int val = 0; if (l1) { val += l1->val; } if (l2) { val += l2->val; } val += carry; carry = val / 10; val = val % 10; node = new ListNode(val); if (prev) { prev->next = node; } addIter(l1 ? l1->next : NULL, l2 ? l2->next : NULL, node, carry); } ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { ListNode *ret = new ListNode(0); addIter(l1, l2, ret, 0); return ret->next; }