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
题解
两个指针,向后扫描,注意进位,当到了最后一个节点时,若进位等于1,则应该创建新节点,并赋值为1。 可以使用头结点,以方便while循环规则。
代码
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class {public : ListNode* addTwoNumbers (ListNode* l1, ListNode* l2) { int carry = 0 ; int sum = 0 ; ListNode * res = new ListNode(0 ); ListNode * cur = res; while (l1 || l2 || carry) { sum = carry; carry = 0 ; if (l1) { sum += l1->val; l1 = l1->next; } if (l2) { sum += l2->val; l2 = l2->next; } if (sum > 9 ) { sum -= 10 ; carry = 1 ; } cur->next = new ListNode(sum); cur = cur->next; } ListNode * tmp = res; res = res->next; delete tmp; return res; } };
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