leetcode(2)Add Two Numbers

Problem

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

这种问题思路上比较简单，不需要考虑如何进行优化，考察点就是代码能力，要做的就是优雅地写出代码。

my solution

就是按照正常竖式手算加法计算两个数中的最短长度，当两个数位数不一样时，在使用多位数加个位数方法。

class Solution(object):    #多位数加个位数    def addListAndNum(self, l, flag):        tmp = l        while l != None:            if l.val + flag >= 10:                l.val = l.val + flag - 10                flag = 1            else:                l.val = l.val + flag                flag = 0                return tmp            pre = l            l = l.next        if flag == 1:            pre.next = ListNode(1)        return tmp    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        tmp = l1        flag = 0        while l1 != None and l2 != None:            if l1.val + l2.val + flag >= 10:                #要注意使用变量与赋值的顺序                l1.val = l1.val + l2.val + flag - 10                flag = 1            else:                l1.val = l1.val + l2.val + flag                flag = 0            p1 = l1            p2 = l2            l1 = l1.next            l2 = l2.next        if l1 == None and l2 == None:            if flag == 1:                p1.next = ListNode(1)        elif l1 == None:            p1.next = self.addListAndNum(l2, flag)        else:            p1.next = self.addListAndNum(l1, flag)        return tmp

时间复杂度：O(max(m,n))
空间复杂度：O(1)

leetcode solution

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {    ListNode dummyHead = new ListNode(0);    ListNode p = l1, q = l2, curr = dummyHead;    int carry = 0;    while (p != null || q != null) {        int x = (p != null) ? p.val : 0;        int y = (q != null) ? q.val : 0;        int sum = carry + x + y;        carry = sum / 10;        curr.next = new ListNode(sum % 10);        curr = curr.next;        if (p != null) p = p.next;        if (q != null) q = q.next;    }    if (carry > 0) {        curr.next = new ListNode(carry);    }    return dummyHead.next;}

对应的python代码：

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        ans = ListNode(0)        tmp = ans        flag = 0        while l1 != None or l2 != None:            if l1 != None:                val1 = l1.val            else:                val1 = 0            if l2 != None:                val2 = l2.val            else:                val2 = 0            if val1 + val2 + flag >= 10:                ans.val = val1 + val2 + flag - 10                flag = 1            else:                ans.val = val1 + val2 + flag                flag = 0            if l1 == None:                pass            else:                    l1 = l1.next            if l2 == None:                pass            else:                l2 = l2.next            ans.next = ListNode(0)            p = ans            ans = ans.next        if flag == 1:            ans.val = 1        else:            p.next = None        return tmp

时间复杂度：O(max(m,n))
空间复杂度：O(max(m,n))

总结

leetcode官方给出的算法的缺点就是空间复杂度还有改进的空间，也可以减小为O(1)的，在我的解法里可以仿照答案，通过一些条件判断把高位空位看做0，可以不使用长整数加个位数的算法。