23

文章讲解
力扣地址

C++

class Solution {
public:
    void reverseString(vector<char>& s) {
        int left = 0;
        int right = s.size() - 1; // right 应该初始化为 s.size() - 1
        while (left < right) {
            swap(s[left], s[right]); // 直接交换 s[left] 和 s[right] 的值
            left++;
            right--;
        }
    }
};

Python

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

---

24

文章讲解
力扣地址

class Solution {
public:
    string reverseStr(string s, int k) {
        int n = s.size();
        for (int i = 0; i < n; i += 2*k) {//每次处理2k个字符
            int left = i;
            int right = min(i + k - 1, n - 1);
            while (left < right) {
                swap(s[left], s[right]);
                left++;
                right--;  // 更新 left 和 right 的值
            }
        }
        return s;
    }
};

class Solution(object):
    def reverseStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        s1=list(s)
        left=0
        right=0
        n = len(s1)
        for i in range(0,n,2*k):
            left = i
            right=min(i+k-1,n-1)
            while left<right:
                s1[left],s1[right]=s1[right],s1[left]
                left+=1
                right-=1
        return "".join(s1)

---

25：

文章讲解
题目地址

#include <iostream>
#include <string>
using namespace std;

int main()
{
    string s;
    while(getline(cin, s)){
        string result = ""; // 用于存储替换后的字符串
        for(int i = 0; i < s.length(); i++){
            if(isdigit(s[i])){ // 判断字符是否为数字
                result += "number";
            } else {
                result += s[i];
            }
        }
        cout << result << endl;
    }
    return 0;
}

判断是不是数字 isdigit(s[i)

while True:
    try:
        
        s = input()
        s1 = list(s)
        for i in range(len(s1)):
            if s1[i]>='0' and s1[i]<='9':
                s1[i]="number"
        print(''.join(s1))
        
    except:
        break

这里注意 ’ '.join(s1)

---

26

文章讲解
题目地址

先翻转字符串再翻转每个单词

然后删除多余空格

reverse(s.begin(),s.end());

if(s[start]!=’ ‘)如果不是空格，说明新单词开始
if(idx!=0) s[idx++]=’ ’ 如果新单词不是第一个单词，加入前单词末尾空格

int end = start;
while(wend<n && s[end]!=‘’)
s[idx++]=s[end++]
通过一个while循环，找到完整的单词

reverse(s.begin()+idx-(end-start),s.begin()=idx);翻转单词

class Solution {
public:
    string reverseWords(string s) {
        // 反转整个字符串
        reverse(s.begin(), s.end());

        int n = s.size();
        int idx = 0;
        for (int start = 0; start < n; ++start) {
            if (s[start] != ' ') {
                // 填一个空白字符然后将idx移动到下一个单词的开头位置
                if (idx != 0) s[idx++] = ' ';

                // 循环遍历至单词的末尾
                int end = start;
                while (end < n && s[end] != ' ') s[idx++] = s[end++];

                // 反转整个单词
                reverse(s.begin() + idx - (end - start), s.begin() + idx);

                // 更新start，去找下一个单词
                start = end;
            }
        }
        s.erase(s.begin() + idx, s.end());
        return s;
    }
};

class Solution(object):
    def reverseWords(self, s):
        """
        :type s: str
        :rtype: str
        """
        words = s.split()
        words = words[::-1]
        result = ' '.join(words)
        return result