# 20. 有效的括号
class Solution {public boolean isValid(String s) {char[] s_Array = s.toCharArray();Stack<Character> stack = new Stack<Character>();for(char c:s_Array){if(c == '('){stack.push(')');}else if(c == '['){stack.push(']');}else if(c == '{'){stack.push('}');}else{if(stack.isEmpty() || stack.pop() != c){return false;}}}if(!stack.isEmpty()){return false;}return true;}
}
1047. 删除字符串中的所有相邻重复项
class Solution {public String removeDuplicates(String s) {Stack<Character> stack = new Stack<Character>();for(char c:s.toCharArray()){if(!stack.isEmpty() && stack.peek() == c){stack.pop();}else{stack.push(c);}}StringBuilder str = new StringBuilder();while(!stack.isEmpty()){str.insert(0,stack.pop());
//这里将弹出的数直接加到字符串的头，就不用再反转字符串return了}return str.toString();}
}
150.逆波特兰表达式求值
这道题几个注意点都在注释中标注
class Solution {public int evalRPN(String[] tokens) {Stack<Integer> stack = new Stack<Integer>();for(String s:tokens){if(isNumber(s)){stack.push(Integer.parseInt(s));}else{if(s.equals("+")){//务必要记住引用数据类型判断相等用equals，因为==比较的是对象的引用，equals是比较的内容stack.push(stack.pop()+stack.pop());}else if(s.equals("/")){//注意这里的顺序，先弹出的作为除数int temp1 = stack.pop();int temp2 = stack.pop();stack.push(temp2/temp1);}else if(s.equals("*")){stack.push(stack.pop() * stack.pop());}else if(s.equals("-")){//注意这里的顺序，先弹出的作为减数int temp1 = stack.pop();int temp2 = stack.pop();stack.push(temp2 - temp1);}}}return stack.pop();}//判断字符串是不是数字public static boolean isNumber(String s){//这里注意判断的应该是最后一个字符，是数字的字符串最后一个字符一定是数字return Character.isDigit(s.toCharArray()[s.length()-1]);}
}