代码随想录二刷day11

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前言

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一、力扣20. 有效的括号

class Solution {public boolean isValid(String s) {Deque<Character> deq = new ArrayDeque<>();for(int i = 0; i < s.length(); i ++){if(s.charAt(i) == '(' || s.charAt(i) == '[' || s.charAt(i) == '{'){deq.offerLast(s.charAt(i));}else{if(deq.isEmpty()){return false;}char temp = deq.pollLast();if(s.charAt(i) == ')' && temp != '('){return false;}if(s.charAt(i) == ']' && temp != '['){return false;}if(s.charAt(i) == '}' && temp != '{'){return false;}}}return deq.isEmpty();}
}

二、力扣1047. 删除字符串中的所有相邻重复项

class Solution {public String removeDuplicates(String s) {Deque<Character> deq = new ArrayDeque<>();StringBuilder sb = new StringBuilder();for(int i = 0; i < s.length(); i ++){if(deq.isEmpty()){deq.offerFirst(s.charAt(i));}else{if(deq.peekFirst() == s.charAt(i)){deq.pollFirst();}else{deq.offerFirst(s.charAt(i));}}}while(!deq.isEmpty()){sb.append(deq.pollLast());}return sb.toString();}
}

三、力扣150. 逆波兰表达式求值

class Solution {public int evalRPN(String[] tokens) {Deque<Integer> deq = new ArrayDeque<>();for(int i = 0; i < tokens.length; i ++){if("+".equals(tokens[i])){int right = deq.pollFirst();int left = deq.pollFirst();deq.offerFirst(left + right);}else if("-".equals(tokens[i])){int right = deq.pollFirst();int left = deq.pollFirst();deq.offerFirst(left - right);}else if("*".equals(tokens[i])){int right = deq.pollFirst();int left = deq.pollFirst();deq.offerFirst(left * right);}else if("/".equals(tokens[i])){int right = deq.pollFirst();int left = deq.pollFirst();deq.offerFirst(left / right);}else{deq.offerFirst(Integer.parseInt(tokens[i]));}}return deq.pollFirst();}
}