LeetCode开心刷题四十五天——93. Restore IP Addresses 92. Reverse Linked List II

93. Restore IP Addresses

Medium

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Given a string containing only digits, restore it by returning all possible valid IP address combinations.

Example:

Input: "25525511135"
Output: ["255.255.11.135", "255.255.111.35"]
BONUS:
1.Python中字符串比较大小，是迭代对比的，先比较第一个元素，看谁大就是谁
如果一样再比下一个，所以35比255大，但是1255比255小.如果想比较字符串数值大小
应该转化为int在其前方调用int()函数
2.地址不能以0开头比如00，02所以如果当前的前一个出现了0要直接舍弃

class Solution(object):
    def restoreIpAddresses(self, s):
        """
        :type s: str
        :rtype: List[str]
        """
        def dfs(s,sub,ips,ip):
            if sub==4:
                if s=='':
                    ips.append(ip[1:])
                return

            for i in range(1,4):
                if i<=len(s):
                    if int(s[:i])<=255:
                        dfs(s[i:],sub+1,ips,ip+'.'+s[:i])
                    if s[0]=='0':break
        ips=[]
        dfs(s,0,ips,'')
        return ips




solu=Solution()
s="25525511135"
print(solu.restoreIpAddresses(s))

92. Reverse Linked List II

Medium

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Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

import collections
# from collections import Counter
# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def reverseBetween(self,head,m,n):
        if head==None or head.next==None:return head
        # 遍历链表的题目中，虚拟节点是必须的，这样最后移动结束才能由它返回起始值
        dummy=ListNode(0)
        dummy.next=head
        head1=dummy
        # 遍历下标才是这种范围的range,计数就直接写一个迭代次数值
        # 不同题目给的数字不同，有的需要减一才是下标有的可以直接来，这里需要减一
        for i in range(m-1):
            head1=head1.next
        p=head1.next
        for i in range(n-m):
            # 链表题就是费指针，这个的核心转换思路是，把p开始指定后，每次p都会向后移位，head直接跟p后面的，p跟下下一个，head的下下一个是开始时，head后面的值
            tmp=head1.next
            head1.next=p.next
            p.next=p.next.next
            head1.next.next=tmp
        return dummy.next
            



solu=Solution()
head=ListNode(1)
l2=ListNode(2)
l3=ListNode(3)
l4=ListNode(4)
l5=ListNode(5)
# l6=ListNode(2)
head.next=l2
l2.next=l3
l3.next=l4
l4.next=l5
# l5.next=l6
m=2
n=4
ans=solu.reverseBetween(head,m,n)
while True:
    print(ans.val)
    if not ans.next:
        break
    ans=ans.next