题目原文:
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
题意解析:
给定两个链表表示两个非负数。数字逆序存储,每个节点包含一个单一的数字。计算两个链表表示的数的和,并以同样格式的链表的形式返还结果。
解法就是直接操作链表相加就可以了。。
代码如下:
C++
class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { ListNode * ans = NULL, *last = NULL; int up = 0; while (NULL != l1 && NULL != l2) { int tmp = l1->val + l2->val + up; up = tmp / 10; if (NULL == last) { ans = new ListNode(tmp % 10); last = ans; } else last = pushBack(last, tmp % 10); l1 = l1->next; l2 = l2->next; } while (NULL != l1) { int tmp = l1->val + up; last = pushBack(last, tmp % 10); up = tmp / 10; l1 = l1->next; } while (NULL != l2) { int tmp = l2->val + up; last = pushBack(last, tmp % 10); up = tmp / 10; l2 = l2->next; } if (0 != up) { ListNode * l = new ListNode(up); last->next = l; } return ans; } ListNode * pushBack(ListNode * last, int val) { ListNode * l = new ListNode(val); last->next = l; return l; } };
Python
# class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @return a ListNode def addTwoNumbers(self, l1, l2): carry = 0; head = ListNode(0); curr = head; while l1 and l2: Sum = l1.val + l2.val + carry carry = Sum / 10 curr.next = ListNode(Sum % 10) l1 = l1.next; l2 = l2.next; curr = curr.next while l1: Sum = l1.val + carry carry = Sum / 10 curr.next = ListNode(Sum % 10) l1 = l1.next; curr = curr.next while l2: Sum = l2.val + carry carry = Sum / 10 curr.next = ListNode(Sum % 10) l2 = l2.next; curr = curr.next if carry > 0: curr.next = ListNode(carry) return head.next
水一枚。。。。
LeetCode第四题,Add Two Numbers
时间: 2024-10-13 15:31:46