上一篇:1.TwoSum(两数之和)

题目(Medium)

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 -> 8
Explanation: 342 + 465 = 807.

给定两个非空链表来代表两个非负数，位数按照逆序方式存储，它们的每个节点只存储单个数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头

脑洞时间

这个题其实求多位数相加，主要的难点就是进位的记录。还有要注意的有：
1.当两个链表长度不一致时的遍历策略，要注意较长的链表的值相当于原值转移
2.当个链表都遍历相加后依然有进位，eg:13+89=102，存储为[3->1],[9->8],遍历相加后为:[2->0],此时8+1+1(上次进位)又形成了进位，需要新增一位[2->0->1](形成的进位)]

废话少说撸代码

# Definition for singly-linked list.
class ListNode:

    def __init__(self, x):
        self.val = x
        self.next = None

class LinkList:
    def __init__(self):
        self.head = None
        self.size = 0

    def isEmpty(self):
        return self.head == None

    def length(self):

        current = self.head
        count = 0

        while current != None:
            count += 1
            current = current.next

        return count

    def append(self, val):
        temp = ListNode(val)

        if self.isEmpty():
            self.head = temp

        else:
            current = self.head
            while current.next != None:
                current = current.next

            current.next = temp

    def add(self, val):
        temp = ListNode(val)

        if self.isEmpty():
            self.head = temp

        else:
            temp.next = self.head
            self.head = temp

        # self.printf()

    def printf(self):
        if self.isEmpty():
            print("NONE")

        else:
            current = self.head
            while current != None:
                print(current.val)
                current = current.next


class Solution:

    # 求当前链表长度
    def length(self, l):
        current = l
        count = 0

        while current != None:
            count += 1
            current = current.next

        return count

    def addTwoNumbers(self, l1, l2):

        rList = LinkList()

        if self.length(l1) >= self.length(l2):
            current1 = l1
            current2 = l2
            count = self.length(l2)

        else:
            current2 = l1
            current1 = l2
            count = self.length(l1)

        zCount = 0
        addFlag = 0

        while current1 != None:
            #按长表来遍历
            tempVal = current1.val
            # 判断短表是否结束
            if zCount < count:
                tempVal = tempVal + current2.val
                current2 = current2.next

            # 判断进位
            if addFlag > 0:
                tempVal = tempVal + addFlag
                addFlag = 0

            # 记录进位
            if tempVal >= 10:
                tempVal = tempVal - 10
                addFlag = 1

            rList.append(tempVal)

            current1 = current1.next
            zCount = zCount + 1

        # 链表都遍历相加完后依然有进位，新增一位
        if addFlag > 0:
            rList.append(addFlag)

        return rList


def main():
    ll1 = LinkList()
    ll2 = LinkList()

    # 342+465=807
    ll1.add(3)
    ll1.add(4)
    ll1.add(2)

    ll2.add(4)
    ll2.add(6)
    ll2.add(5)
    # ll2.add(0)

    # l1.printf()
    # l2.printf()

    l1 = ll1.head
    l2 = ll2.head

    rl = Solution().addTwoNumbers(l1, l2)
    rl.printf()

if __name__ == '__main__':
    main()

算法复杂度

时间复杂度为:O(N)，N为较长链表的长度，即

空间复杂度为:O(N)

三省吾题

刚开始以为操作的是两个链表，写参数的时候就传了两个链表。Leetcode运行老是报错，说有些方法没有定义。仔细一看，才发现传的参数是节点指针，稍作修改乎就Accepted了。

总结而言，这道题并不难，主要考察数据结构中链表的构造与遍历思路很快就有了，我花费的时间有点长是因为在顺便练习Python的语法。可不要笑话我这个Python新手……

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下一篇:3.无重复字符的最长子串(LongestSubstringWithoutRepeatingCharacters)

——————————20180311夜

刷Leetcode，

在我看来，

其实是为了维持一种编程状态！