LeetCode-addTwoNumbers-python

Description

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

Solution

时间复杂度：O(max(m, n))

空间复杂度：O(max(m, n))

核心：
注意链表长度不一致和进位即可。

1. 由于链表长度可能不一致，因此只要任一链表有值或有carray（进位），就继续计算
2. 循环中判断，链表如果无值那么就取0，并且每个链表走向下一个节点

def addTwoNumbers(self, l1, l2):
    ret = dummy = ListNode(0)
    carry = 0
    while l1 or l2 or carry:
        v1 = l1.val if l1 else 0
        v2 = l2.val if l2 else 0
        v_sum = v1 + v2 + carry
        if (v_sum >= 10):
            carry = 1
            v_sum = v_sum - 10
        else:
            carry = 0
        dummy.next = ListNode(v_sum)
        dummy = dummy.next
        l1 = l1.next if l1 else 0
        l2 = l2.next if l2 else 0
    return ret.next

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