✅DAY8 字符串str | 344.反转字符串 | 541. 反转字符串II | 卡码网：54.替换数字

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解题思路：双指针解法，直接把两边调转

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

解题思路：reversed

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        s[:] = reversed(s)

541. 反转字符串 II

解题思路：灵活应用左右指针，每次loop的step = 2*k步。同时把str改为list更好替换

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

        for char in range(0, len(s), 2*k):
            res[char: char+k] = reverse_substring(res[char: char+k])

        return ''.join(res)

解题思路：reverse可以直接用text[::-1]。 定义需要reverse和keep的部分，累加到res

class Solution:
    def reverseStr(self, s: str, k: int) -> str:

        def reverse_substring(text):
            return text[::-1]  # 使用切片来反转字符串

        res = ""

        # 按照每段 2*k 的长度进行遍历
        for i in range(0, len(s), 2 * k):
            # 对于每一段，前 k 个字符反转，后面的保持不变
            res += reverse_substring(s[i:i + k]) + s[i + k:i + 2 * k]
        
        return res

卡码网：54.替换数字

解题思路：python解法好像没那么多弯弯绕绕

class Solution:
    def change(self, s: str) -> str:
        res = ""
        for char in s:
            if char.isdigit():
                res += "number"  # 拼接 "number"
            else:
                res += char
        return res

更简单一点的写法

class Solution:
    def change(self, s: str) -> str:
        return ''.join("number" if char.isdigit() else char for char in s)