Python leetcode #2 Add Two Numbers
2017-05-01 20:14
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leetcode #2 Add Two Numbers
题我就不读了,来讲下思路。1. 注意每个节点有两个属性,一个本身的值,另一个指向下一个节点。
2. 然后建立一个虚拟的头节点,它的值为0,这个节点最后不存在,只是暂时拿来利用。
3. 把头结点赋给当前节点cur,把l1, l2赋给类内变量,设立进位carry的值为0.
4. 建立循环测试l1, l2是否有值,如果没值的话说明已经空了。和等于进位加上两个节点当前位的值,进位的值等住整除的值(非0即1),当前节点的下一节点的值等于和对10取余。把当前节点的下一个节点赋给当前节点。测试l1, l2的下一个节点是否有值,有值赋给自己没值赋空。
5. 循环完后检查最后一次进位,如有值把他赋给当前节点的下一节点。
6. 返回虚拟头节点的下一节点
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ head = ListNode(0) cur = head carry = 0 p = l1 q = l2 while p is not None or q is not None: x = p.val if p is not None else 0 y = q.val if q is not None else 0 sum = carry + x + y carry = sum // 10 cur.next = ListNode(sum % 10) cur = cur.next p = p.next if p is not None else None q = q.next if q is not None else None if carry > 0: cur.next = ListNode(carry) return head.next
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