2. 两数相加 [medium]

2020-05-31

https://leetcode-cn.com/problems/add-two-numbers/

second try

执行用时 :76 ms, 在所有 Python 提交中击败了53.08%的用户

内存消耗 :12.9 MB, 在所有 Python 提交中击败了5.55%的用户

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

Solution(object):
    def addTwoNumbers(self, l1, l2):
        rv = node = ListNode(None)
        carry = 0
        while(l1 or l2 or carry):
            left = l1.val if l1 else 0
            right = l2.val if l2 else 0
            s = left + right + carry
            carry, val = divmod(s, 10)
            node.next = ListNode(val)
            node = node.next
            l1 = l1.next if l1 else 0
            l2 = l2.next if l2 else 0
        return rv.next   

first try

执行用时 :112 ms, 在所有 Python 提交中击败了11.05%的用户

内存消耗 :12.7 MB, 在所有 Python 提交中击败了5.55%的用户

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        print( self.origianl_val(l1), self.origianl_val(l2))
        s = self.origianl_val(l1) + self.origianl_val(l2)
        return self.val_to_link(s)

    def origianl_val(self,link):
        rv = link.val
        power = 1
        while(link.next):
            link = link.next
            rv +=  link.val * (10 ** power)
            power += 1
        return rv

    def val_to_link(self, val):
        rv = node = ListNode()
        q, r = divmod(val, 10)
        while(q != 0):
            node.val = r
            node.next = ListNode()
            node = node.next
            val = (val - r) / 10
            q, r = divmod(val, 10)
        node.val = r
        return rv

2. 两数相加 [medium]​

2. 两数相加 [medium]