代码随想录训练营二刷第十一天 | 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

代码随想录训练营二刷第十一天 | 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

一、20. 有效的括号

题目链接：https://leetcode.cn/problems/valid-parentheses/
思路：思路遇到左括号把对应的右括号压入栈，节省后续判断。

class Solution {
    public static boolean isValid(String s) {
        LinkedList<Character> stack = new LinkedList<>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '[') {
                stack.push(']');
            }else if (s.charAt(i) == '{') {
                stack.push('}');
            }else if (s.charAt(i) == '(') {
                stack.push(')');
            }else if (stack.isEmpty() || stack.peek() != s.charAt(i)) {
                return false;
            }else {
                stack.pop();
            }
        }
        return stack.isEmpty();
    }
}

二、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/
思路：简单的匹配

public String removeDuplicates(String s) {
        LinkedList<Character> stack = new LinkedList<>();
        for (int i = 0; i < s.length(); i++) {
            if (stack.isEmpty() || stack.peek() != s.charAt(i)) {
                stack.push(s.charAt(i));
            }else {
                stack.pop();
            }
        }
        StringBuilder result = new StringBuilder();
        while (!stack.isEmpty()) {
            result.insert(0, stack.pop());
        }
        return result.toString();
    }

三、150. 逆波兰表达式求值

题目链接：https://leetcode.cn/problems/evaluate-reverse-polish-notation/
思路：每次出栈，出两个运算再压栈。

class Solution {
    public int evalRPN(String[] tokens) {
        LinkedList<Integer> stack = new LinkedList<>();
        for (int i = 0; i < tokens.length; i++) {
            if ("+".equals(tokens[i])) {
                int temp = stack.pop();
                stack.push(stack.pop() + temp);
            }else if ("-".equals(tokens[i])) {
                int temp = stack.pop();
                stack.push(stack.pop() - temp);
            }else if ("*".equals(tokens[i])) {
                int temp = stack.pop();
                stack.push(stack.pop() * temp);
            }else if ("/".equals(tokens[i])) {
                int temp = stack.pop();
                stack.push(stack.pop() / temp);
            }else {
                stack.push(Integer.valueOf(tokens[i]));
            }
        }
        return stack.pop();
    }
}