回文字符串判断汇总

Lc 125 验证回文字符串

给定一个字符串，验证它是否是回文串，只考虑字母和数字字符，可以忽略字母的大小写。

class Solution:
    def judgeNumOrStr(self,a):
        if ord('0')<=ord(a)<=ord('9') or ord('a')<=ord(a)<=ord('z'):
            return True
        else:
            return False

    def isPalindrome(self, s: str) -> bool:
        s = s.lower()

        i,j = 0,len(s)-1
        while i<j:
            if not self.judgeNumOrStr(s[i]):
                i+=1
                continue
            if not self.judgeNumOrStr(s[j]):
                j-=1
                continue
            if s[i] != s[j]:
                return False
            i+=1
            j-=1
        return True

 

Lc 680 验证回文字符串Ⅱ

给定一个非空字符串 s，最多删除一个字符。判断是否能成为回文字符串。

先写一个单独验证回文字符串的函数A，然后判断的时候如果不相等则调用A判断 [i+1,j] 或者 [i,j-1]是否为回文

class Solution:
    def judgestr(self,s,i,j):
        while i<j:
            if s[i] != s[j]:
                return False
            i+=1
            j-=1
        return True
        
    def validPalindrome(self, s: str) -> bool:
        i, j = 0,len(s)-1
        count = 0
        while i<j:
            if s[i] != s[j]:
                if self.judgestr(s,i+1,j) or self.judgestr(s,i,j-1):
                    return True
                else:
                    return False
            i+=1
            j-=1
                
        return True

 

Lc 234 回文链表

请判断一个链表是否为回文链表。

用 O(n) 时间复杂度和 O(1) 空间复杂度解决

用快慢指针找到中间节点，反转后半部分，然后进行对比

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

class Solution:
    def reverseList(self,head):
        pre = None
        p = head
        while p:
            nextNode = p.next
            p.next = pre
            pre = p
            p = nextNode
        return pre

    def isPalindrome(self, head: ListNode) -> bool:
        if not head or not head.next:
            return True
        slow,fast = head,head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
        
        mid = slow.next
        slow.next = None
        left = head
        right = self.reverseList(mid)

        while left and right:
            if left.val != right.val:
                return False
            left = left.next
            right = right.next
        return True

 

Lc 5 最长回文子串

给定一个字符串 s，找到 s 中最长的回文子串。你可以假设 s 的最大长度为 1000。

动态规划：

判断s[i,j]时，需要知道s[i+1,j-1]的状态，因此遍历时i从大到小(n-1->0),j从小到大(i->n-1)

class Solution:
    def longestPalindrome(self, s: str) -> str:
        if not s or len(s) == 0:
            return ""
        n,start,maxlen = len(s),0,0

        d = [[False]*n for _ in range(n)]
         
        for i in range(n-1,-1,-1):
            for j in range(i,n):
                if i==j:
                    d[i][j] = True
                elif i+1==j:
                    d[i][j] = (s[j] == s[i])
                else:
                    d[i][j] = ((s[i]==s[j]) and d[i+1][j-1])
            
                if d[i][j] and j-i+1>maxlen:
                    maxlen = (j-i+1)
                    start = i
                
        return s[start:start+maxlen]