Leetcode刷题笔记1 字符串part02

leetcode151 翻转字符串中的单词

不适用split库函数：

class Solution:
    def reverseWords(self, s: str) -> str:
        s = s.strip()
        res = []
        l, r = len(s) - 1,len(s) - 1
        while l >= 0:
            while l >= 0 and s[l] != ' ':
                l -= 1
            res.append(s[l + 1:r + 1])
            while l >= 0 and s[l] == ' ':
                l -= 1
                r = l
        return ' '.join(res)
        

使用split库函数：

class Solution:
    def reverseWords(self, s: str) -> str:
        s = s[::-1]
        return ' '.join(i[::-1] for i in s.split())

卡码网 55 右旋字符串

k = int(input())
s = input()

print(s[-k:] + s[:-k])

leetcode28 找出字符串中第一个匹配项的下标

kmp算法入门
首先要明白什么是子串，比如aabaaf 的子串是a、aa、aab、aaba、aabaa、aabaaf

其次寻找最长相等前后缀，即get_next数组。

class Solution:
    # 核心技术得到next数组
    def get_next(self, s):
        j, prefix = 0, [0]*len(s)
        for i in range(1, len(s)):
            while j > 0 and s[j] != s[i]:
                j = prefix[j - 1]
            if s[j] == s[i]:
                j += 1
            prefix[i] = j
        return prefix
    def strStr(self, haystack: str, needle: str):
        j, prefix = 0, self.get_next(needle)
        for i in range(len(haystack)):
            while j > 0 and haystack[i] != needle[j]:
                j = prefix[j - 1]
            if haystack[i] == needle[j]:
                j += 1
            if j == len(needle):
                return i - j + 1
        return -1

leetcode 459 重复的子字符串

思路：如果字符串s 在s+s[1:-1]中可以找到,那么它由子串构成

class Solution:
    def repeatedSubstringPattern(self, s: str) -> bool:
        if s in (s + s)[1:-1]:
            return True
        else:
            return False

kmp算法思路：如果由重复的子字符串构成，则len（s）- next[-1]可以被len（s）整除且next[-1]不为0

class Solution:
    def repeatedSubstringPattern(self, s: str) -> bool:  
        if len(s) == 0:
            return False
        nxt = [0] * len(s)
        self.getNext(nxt, s)
        if nxt[-1] != 0 and len(s) % (len(s) - nxt[-1]) == 0:
            return True
        return False
    
    def getNext(self, nxt, s):
        nxt[0] = 0
        j = 0
        for i in range(1, len(s)):
            while j > 0 and s[i] != s[j]:
                j = nxt[j - 1]
            if s[i] == s[j]:
                j += 1
            nxt[i] = j
        return nxt