class Solution { public void reverseString(char[] s) { } }把代码改成这个模板

可以按照以下方式将你的代码改写成题目要求的模板：

class Solution {
    public void reverseString(char[] s) {
        int left = 0, right = s.length - 1;
        while (left < right) {
            char temp = s[left];
            s[left++] = s[right];
            s[right--] = temp;
        }
    }
}

在这个模板中，我们将你所提供的reverseString()方法修改为了符合题目要求的形式。函数的参数为一个字符数组s，函数返回值为void。在函数内部，我们使用了双指针的方法对字符数组s进行了原地翻转，将字符串反转。

相关问题

对class Solution{ public: string removeDuplicateLetters(string s) { vector<int> count(26, 0); vector<bool> inStack(26, false); stack<char> st; for (char c : s) { count[c - 'a']++; } for (char c : s) { count[c - 'a']--; if (inStack[c - 'a']) { continue; } while (!st.empty() && st.top() > c && count[st.top() - 'a'] > 0) { inStack[st.top() - 'a'] = false; st.pop(); } st.push(c); inStack[c - 'a'] = true; } string ans = ""; while (!st.empty()) { ans += st.top(); st.pop(); } reverse(ans.begin(), ans.end()); return ans; } };进行封装，涉及基本数据结构和容器的，需自行编写相应的类，禁止直接使用STL的数据容器类。

好的，对于该题解的代码进行封装的话，我们可以自己实现栈和向量容器类，也可以使用已有的容器类进行封装。以下是使用已有的容器类进行封装的代码：

#include <string>
#include <vector>
#include <stack>
#include <algorithm>

class Solution{
public:
    std::string removeDuplicateLetters(std::string s){
        MyVector<int> count(26, 0);
        MyVector<bool> inStack(26, false);
        MyStack<char> st;
        for (char c : s){count[c - 'a']++;}
        for (char c : s){
            count[c - 'a']--;
            if (inStack[c - 'a']){continue;}
            while (!st.empty() && st.top() > c && count[st.top() - 'a'] > 0){
                inStack[st.top() - 'a'] = false; st.pop();
            }
            st.push(c);
            inStack[c - 'a'] = true;
        }
        std::string ans = "";
        while (!st.empty()){ans += st.top(); st.pop();}
        std::reverse(ans.begin(), ans.end());
        return ans;
    }

private:
    template <typename T>
    class MyVector{
    public:
        MyVector(): size(0), capacity(1), data(new T[capacity]){}
        MyVector(int initCapacity, T initValue): size(initCapacity), capacity(initCapacity), data(new T[capacity]){
            for (int i = 0; i < size; i++){
                data[i] = initValue;
            }
        }
        MyVector(const MyVector& other): size(other.size), capacity(other.capacity), data(new T[capacity]){
            for (int i = 0; i < size; i++){
                data[i] = other.data[i];
            }
        }
        ~MyVector(){delete[] data;}

        void push_back(T value){
            if (size == capacity){
                capacity *= 2;
                T* newData = new T[capacity];
                for (int i = 0; i < size; i++){
                    newData[i] = data[i];
                }
                delete[] data;
                data = newData;
            }
            data[size++] = value;
        }

        bool empty(){return size == 0;}
        int size(){return size;}
        T& operator[](int index){return data[index];}

    private:
        int size;
        int capacity;
        T* data;
    };

    template <typename T>
    class MyStack{
    public:
        MyStack(): size(0), capacity(1), data(new T[capacity]){}
        ~MyStack(){delete[] data;}

        void push(T value){
            if (size == capacity){
                capacity *= 2;
                T* newData = new T[capacity];
                for (int i = 0; i < size; i++){
                    newData[i] = data[i];
                }
                delete[] data;
                data = newData;
            }
            data[size++] = value;
        }

        void pop(){size--;}
        T top(){return data[size - 1];}
        bool empty(){return size == 0;}

    private:
        int size;
        int capacity;
        T* data;
    };
};

在这里，我们使用了模板类来实现自己的向量和栈容器类，这样可以方便地适用于不同类型的数据。同时，我们也将向量和栈容器类的实现放在了类内部，防止了与其他代码的命名空间冲突。

给我写出完整的可运行的代码public class Solution { public String removeDuplicateLetters(String s) { int[] tem = new int[26]; for (int i = 0; i < s.length(); i++) { char tems = s.charAt(i); tem[tems - 'a']++; } boolean[] visited = new boolean[26]; Stack<Character> stack = new Stack<>(); for (int i = 0; i < s.length(); i++) { char get = s.charAt(i); while (!stack.isEmpty() && stack.peek() > get && !visited[get - 'a'] && tem[stack.peek() - 'a'] > 0) { visited[stack.peek() - 'a'] = false; stack.pop(); } if (visited[get - 'a']) { tem[get - 'a']--; continue; } visited[get - 'a'] = true; stack.push(get); tem[get - 'a']--; } StringBuilder sBuilder = new StringBuilder(); while (!stack.isEmpty()) { sBuilder.append(stack.pop()); } return sBuilder.reverse().toString(); }的 }

这里是完整可运行的代码，您可以直接复制粘贴并运行：

import java.util.*;

public class Solution {
    public String removeDuplicateLetters(String s) {
        int[] tem = new int[26];
        for (int i = 0; i < s.length(); i++) {
            char tems = s.charAt(i);
            tem[tems - 'a']++;
        }
        boolean[] visited = new boolean[26];
        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            char get = s.charAt(i);
            while (!stack.isEmpty() && stack.peek() > get && !visited[get - 'a'] && tem[stack.peek() - 'a'] > 0) {
                visited[stack.peek() - 'a'] = false;
                stack.pop();
            }
            if (visited[get - 'a']) {
                tem[get - 'a']--;
                continue;
            }
            visited[get - 'a'] = true;
            stack.push(get);
            tem[get - 'a']--;
        }
        StringBuilder sBuilder = new StringBuilder();
        while (!stack.isEmpty()) {
            sBuilder.append(stack.pop());
        }
        return sBuilder.reverse().toString();
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        String s = "bcabc";
        String result = solution.removeDuplicateLetters(s);
        System.out.println(result);
    }
}

在这个示例中，我们调用了removeDuplicateLetters方法来移除字符串"bcabc"中的重复字符，并将结果打印到控制台上。