题目浅析

  • 想查看原题可以点击题目链接。

  • 简单地说，就是给一个链表，从头到尾就是从大到小表示一个数字的位，现在要把这个数字乘 2，求最终的数字链表。

思路分享

代码解答（强烈建议自行解答后再看）

  • 参考题解
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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head):
        pre = None
        while head:
            nxt = head.next
            head.next = pre
            pre = head
            head = nxt
        return pre

    def doubleIt(self, head: Optional[ListNode]) -> Optional[ListNode]:
        head = self.reverseList(head)
        cur = head
        add_nxt = 0
        while cur:
            cur.val *= 2
            cur.val += add_nxt
            add_nxt = 0
            if cur.val >= 10:
                add_nxt = 1
                cur.val -= 10
            
            if not cur.next:
                break
            cur = cur.next
            
        if add_nxt > 0:
            cur.next = ListNode(1)
        
        return self.reverseList(head)