[LeetCode]Add Two Numbers II

题目描述:

LeetCode 445. Add Two Numbers II

You are given two linked lists representing two non-negative numbers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

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

Output: 7 -> 8 -> 0 -> 7

题目大意:

给定两个表示非负整数的链表。链表头部表示高位,尾部表示低位,每一个节点只包含一位数。以链表形式返回两个链表的和。

你可以假设两个数字都不包含前导0,除非数字本身就是0。

进一步思考:

你可以不修改输入链表吗?换言之,不允许反转链表。

解题思路:

双指针(Two Pointers)

时间复杂度O(n),空间复杂度O(n),所用空间除保存最终结果外,没有额外开销

具体步骤如下:

1. 统计两链表长度s1, s2;最终结果链表长度s = max(s1, s2) (若有进位,则为s+1)

2. 将两链表对齐并逐节点求和,记头节点为h(头节点为dummy node,最高位从h.next开始)

3. 初始令指针p指向头节点h,执行循环:

    令指针q = p.next,重复向其下一节点移动,直到q为空或者q.val ≠ 9
    
    如果q.val > 9,说明p与q之间的所有节点需要进位,令p向q移动同时修改p.val
    
    否则,令p = q

Python代码:

# 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
        """
        s1 = self.getSize(l1)
        s2 = self.getSize(l2)
        s = max(s1, s2)

        p = h = ListNode(0)
        while s:
            p.next = ListNode(0)
            p = p.next
            if s <= s1:
                p.val += l1.val
                l1 = l1.next
            if s <= s2:
                p.val += l2.val
                l2 = l2.next
            s -= 1

        p = h
        while p:
            q = p.next
            while q and q.val == 9:
                q = q.next
            if q and q.val > 9:
                while p != q:
                    p.val += 1
                    p = p.next
                    p.val -= 10
            else: p = q
        return h if h.val else h.next
    
    def getSize(self, h):
        c = 0
        while h:
            c += 1
            h = h.next
        return c

 

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