代码随想录算法训练营第十天-20有效的括号、1047删除字符串中的所有相邻重复项、150逆波兰表达式求值

前言
状态：20AC，1047AC，150AC
更新

1	24.05.30 初始记录

初步题解

20 有效的括号

题目链接：(https://leetcode.cn/problems/valid-parentheses/description/)

public class LE20 {  
    public static void main(String[] args) {  
        Scanner scanner = new Scanner(System.in);  
        System.out.println(isValid(scanner.nextLine()));  
    }  
  
    public static boolean isValid(String s) {  
        Stack<Character> stack = new Stack<>();  
        for (char c : s.toCharArray()) {  
            if (c == '(' || c == '[' || c == '{') {  
                stack.push(c);  
                continue;  
            }  
            if (stack.isEmpty()) {  
                return false;  
            }  
            Character pop = stack.pop();  
            if ((c == ')' && pop != '(' ) || (c == ']' && pop != '[' ) || (c == '}' && pop != '{' )) {  
                return false;  
            }  
        }  
        return stack.isEmpty();  
    }  
}

1047 删除字符串中的所有相邻重复项

题目链接：(https://leetcode.cn/problems/remove-all-adjacent-duplicates-in-string/description/)

public static void main(String[] args) {  
    Scanner scanner = new Scanner(System.in);  
    System.out.println(removeDuplicates(scanner.nextLine()));  
}  
  
public static String removeDuplicates(String s) {  
    Stack<Character> stack = new Stack<>();  
    for (char c : s.toCharArray()) {  
        if (stack.isEmpty()) {  
            stack.push(c);  
            continue;  
        }  
        if (c == stack.peek()) {  
            stack.pop();  
        } else {  
            stack.push(c);  
        }  
    }  
  
    StringBuilder str = new StringBuilder();  
    int size = stack.size();  
    for (int i = 0; i < size; i++) {  
        str.insert(0, stack.pop());  
    }  
    return str.toString();  
}

150 逆波兰表达式求值

题目链接：(https://leetcode.cn/problems/evaluate-reverse-polish-notation/description/)

可以直接根据题目中给的链接查看逆波兰表示法的意思

public int evalRPN(String[] tokens) {  
    Stack<Integer> stack = new Stack<>();  
    int num = Integer.parseInt(tokens[0]);  
    for (String token : tokens) {  
        switch (token) {  
            case "+": {  
                num = stack.pop() + stack.pop();  
                stack.push(num);  
                break;  
            }  
            case "-": {  
                Integer num2 = stack.pop();  
                Integer num1 = stack.pop();  
                num = num1 - num2;  
                stack.push(num);  
                break;  
            }  
            case "*": {  
                num = stack.pop() * stack.pop();  
                stack.push(num);  
                break;  
            }  
            case "/": {  
                Integer num2 = stack.pop();  
                Integer num1 = stack.pop();  
                num = num1 / num2;  
                stack.push(num);  
                break;  
            }  
            default: {  
                stack.push(Integer.parseInt(token));  
            }  
        }  
    }  
    return num;  
}

看解析