用链表表示的两个数相加
2015-08-18 22:03
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1 题目
You are giventwo linked lists representing two non-negative numbers. The digits are storedin reverse order and each of their nodes contain a single digit. Add the twonumbers and return it as a linked list.Input: (2 -> 4 -> 3)+ (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
2 分析
该题目属于链表的相加,需要注意的有:(1) 考虑两个链表的长度,尤其是链表为空时也能处理。
(2) 每个结点只表示一位数字。
(3) 当链表末尾结点相加后若有进位,则需要申请新的结点存储信息。
3 实现
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ intlistLength(ListNode* head) { return head ? 1 + listLength(head ->next) : 0; } ListNode*Solution::addTwoNumbers(ListNode *l1, ListNode *l2) { if (listLength(l1) < listLength(l2)) { return addTwoNumbers(l2, l1); } ListNode *head1 = l1, *head2 = l2; int inc = 0; bool isEnd = false; while (head2) { int val = head1 -> val + head2-> val + inc; head1 -> val = val % 10; inc = val / 10; if (head1 -> next) { head1 = head1 -> next; } else { isEnd = true; } head2 = head2 -> next; } while (inc) { int val = isEnd ? inc : head1 ->val + inc; if (isEnd) { head1 -> next = newListNode(val % 10); } else { head1 -> val = val % 10; } inc = val / 10; if (head1 -> next) { head1 = head1 -> next; } else { isEnd = true; } } return l1; }
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