LeetCode33 - 技术栈

445.两数相加Ⅱ

给你两个非空链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。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。

示例1：

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输入：l1 = [7,2,4,3], l2 = [5,6,4]
输出：[7,8,0,7]

示例2：

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输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[8,0,7]

示例3：

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输入：l1 = [0], l2 = [0]
输出：[0]

提示：

链表的长度范围为 [1, 100]
0 <= node.val <= 9
输入数据保证链表代表的数字无前导 0

**进阶：**如果输入链表不能翻转该如何解决？

python 复制代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        p, q = l1, l2
        n1, n2 = 0, 0
        while p:
            p = p.next
            n1 += 1
        while q:
            q = q.next
            n2 += 1
        
        l3 = ListNode()
        p, q, j = l1, l2, l3
        while p and q:
            if n1 > n2:
                x = p.val
                y = 0
                n1 -= 1
                p = p.next
            elif n2 > n1:
                x = 0
                y = q.val
                n2 -= 1
                q = q.next
            else:
                x = p.val
                y = q.val
                n1 -= 1
                n2 -= 1
                p = p.next
                q = q.next
            if x + y >= 10:
                node = ListNode(x+y-10)
                j.val += 1
            else:
                node = ListNode(x+y)
            j.next = node
            j = j.next
        
        pre = None
        cur = l3
        while cur:
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt
        #return pre

        kk = pre
        while kk:
            if kk.val >= 10:
                kk.val -= 10
                if kk.next:
                    kk.next.val += 1
                else:
                    node = ListNode(1)
                    kk.next = node
            kk = kk.next

        pre1 = None
        cur1 = pre
        while cur1:
            nxt1 = cur1.next
            cur1.next = pre1
            pre1 = cur1
            cur1 = nxt1
        
        if pre1.val == 0:
            return pre1.next
        else:
            return pre1

python 复制代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        pre1 = None
        cur = l1
        while cur:
            nxt = cur.next
            cur.next = pre1
            pre1 = cur
            cur = nxt
        #return pre1

        pre2 = None
        cur = l2
        while cur:
            nxt = cur.next
            cur.next = pre2
            pre2 = cur
            cur = nxt
        #return pre2

        cur = l3 = ListNode()
        carry = 0
        while pre1 or pre2 or carry:
            s = carry + (pre1.val if pre1 else 0) + (pre2.val if pre2 else 0)
            cur.next = ListNode(s % 10)
            carry = s // 10
            cur = cur.next
            if pre1:
                 pre1 = pre1.next
            if pre2:
                 pre2 = pre2.next
        #return l3.next

        pre3 = None
        cur = l3.next
        while cur:
            nxt = cur.next
            cur.next = pre3
            pre3 = cur
            cur = nxt
        return pre3

不翻转链表：把单链表先依次储存到栈里，再依次出栈。