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

偏执的太偏执、 2023-08-17 15:15 103阅读 0赞

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

转载于:https://www.cnblogs.com/Marigolci/p/11555419.html