Leetcode 刷題之路 2. Add two numbers

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8Explanation: 342 + 465 = 807.

Repeat question:

有兩個非空鏈表，代表兩個非負整數，數字在鏈表內逆序存儲，求兩個數的和，並用同樣的逆序方式存儲在鏈表中。

Solution：

解法一

設一個函數number(), 將一個鏈錶轉換成整型的數， 用這個函數計算兩個鏈表的整數然後求和。 求和完後，將這個整數變為字元串數組形式，然後將逆序後的數組的每一個元素變為整型後寫入鏈表。

解法二

考慮到number()函數如果計算的數字過大會越界，我們改良number()函數為stringNumber(), 這個函數依次將鏈表中的每一位存入數組。在求和函數中，調用這個函數以此求得兩個數組並逆序，設置flag表示是否有進位，初始值為0. 然後每次彈出兩個數組的整數並相加再加上flag, 直到兩個數組都為空了結束。

Code:

解法一：

# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: def number(self, l): count = 0 result = 0 while l: result += l.val* 10**count l = l.next count +=1 return result def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ result = self.number(l1)+ self.number(l2) result = str(result) head = None for i, res in enumerate(result[::-1]): if i is 0: l = ListNode(int(res)) head = l else: l.next = ListNode(int(res)) l = l.next return head

解法二：

# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: def stringNumber(self, l): result = [] while l: result.append(l.val) l = l.next return result def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ number1 = self.stringNumber(l1)[::-1] number2 = self.stringNumber(l2)[::-1] result = [] flag = 0 while number1 or number2 or flag: if number1 and number2: res = number1.pop() + number2.pop() + flag if res > 9: flag = 1 else: flag = 0 result.append(res%10) elif number1: res = number1.pop() + flag if res > 9: flag = 1 else: flag = 0 result.append(res%10) elif number2: res = number2.pop() + flag if res > 9: flag = 1 else: flag = 0 result.append(res%10) else: if flag: result.append(flag) flag = 0 head = None for i, res in enumerate(result): if i is 0: l = ListNode(int(res)) head = l else: l.next = ListNode(int(res)) l = l.next return head

Test case:

l1: 0->1l2: 0->1—>2l1: 9->9l2: 1l1: 1->2->3........->9l1: 1->2->3....... ->9

Complexity analysis:

Time complexity: O(max(m, n)). assume that m and n represents the length of l1 and l2 respectively, the algorithm above iterates at most max(m, n) times.

Follow up:

what if the digits in the linked list are stored in non-reversed order?

(3 -> 4 -> 2) + (4 -> 6 -> 5) = 8 -> 0 -> 7

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