代码训练营第八天（Python）｜ 344.反转字符串、541. 反转字符串II、05.替换空格、151.翻转字符串里的单词、58-II.左旋转字符串

344.反转字符串

双指针法
时间复杂度为: O(n), 空间复杂度为: O(1)

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        left, right = 0, len(s) - 1
        while left < right:
            s[left], s[right] = s[right], s[left]
            left += 1
            right -= 1
        return s

541. 反转字符串II

注意点: 使用切片处理掉边界的问题
方法一

class Solution:
    def reverseStr(self, s: str, k: int) -> str:
        def helper(s_list:list):
            left, right = 0, len(s_list)-1
            while left < right:
                s_list[left], s_list[right] = s_list[right], s_list[left]
                left += 1
                right -= 1
            return s_list

        res_s = list(s)
        for i in range(0, len(s), 2*k):   # 每次 2k 的步距
            res_s[i:i+k] = helper(res_s[i:i+k]) # 反转 k 或者小于k的列表
        return "".join(res_s)

方法二

class Solution:
    def reverseStr(self, s: str, k: int) -> str:
        start = 0    # start 开始位置
        while start < len(s):
            rev_p = start + k   # rev_p 需要反转的位置
            s = s[:start] + s[start:rev_p][::-1] + s[rev_p:]
            start += 2*k   # 每次 2k 步距移动
        return s

05.替换空格

class Solution:
    def pathEncryption(self, path: str) -> str:
        res = ""
        for i in path:
            if i == ".":
                res += " "
            else:
                res += i
        return res

151.翻转字符串里的单词

class Solution:
    def reverseWords(self, s: str) -> str:
        s_list = s.split() # split 不传参数获取非空格的元素
        left, right = 0, len(s_list) - 1
        while left < right:
            s_list[left], s_list[right] = s_list[right], s_list[left]
            left += 1
            right -= 1
        return " ".join(s_list)

58-II.左旋转字符串

1、使用切片

class Solution:
    def dynamicPassword(self, password: str, target: int) -> str:
        return password[target:] + password[:target]

2、使用自定义的reverse,以列表反转方式代替其他语言的原地字符串操作

class Solution:
    def dynamicPassword(self, password: str, target: int) -> str:
        password_l = list(password)
        self.reverse(password_l, 0, target-1)
        self.reverse(password_l, target, len(password_l)-1)
        self.reverse(password_l, 0, len(password_l)-1)
        return "".join(password_l)


    def reverse(self, s_list, start, end):
        while start < end:
            s_list[start], s_list[end] = s_list[end], s_list[start]
            start += 1
            end -= 1