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![https://leetcode.com/problems/evaluate-reverse-polish-notation/] Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Note: Division between two integers should truncate toward zero. The given RPN expression is always valid. That means the expression would always evaluate to a result and there won’t be any divide by zero operation. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Example 1: Input: ["2", "1", "+", "3", "*"] Output: 9 Explanation: ((2 + 1) * 3) = 9 Example 2: Input: ["4", "13", "5", "/", "+"] Output: 6 Explanation: (4 + (13 / 5)) = 6 Example 3: Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"] Output: 22 Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5 = ((10 * (6 / (12 * -11))) + 17) + 5 = ((10 * (6 / -132)) + 17) + 5 = ((10 * 0) + 17) + 5 = (0 + 17) + 5 = 17 + 5 = 22 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 class Solution { public int evalRPN(String[] tokens) { int[] stack = new int[tokens.length]; int size = 0; for (String token: tokens) { if (token.equals("+") || token.equals("-") || token.equals("*") || token.equals("/") ) { int b = stack[--size]; int a = stack[--size]; int c; switch (token) { case "+": c = a + b; break; case "-": c = a - b; break; case "*": c = a * b; break; case "/": c = a / b; break; default: throw new RuntimeException(); } stack[size++] = c; } else { stack[size++] = Integer.valueOf(token); } } return stack[0]; } }

![https://leetcode.com/problems/crawler-log-folder/] The Leetcode file system keeps a log each time some user performs a change folder operation. The operations are described below: “../” : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder). “./” : Remain in the same folder. “x/” : Move to the child folder named x (This folder is guaranteed to always exist). You are given a list of strings logs where logs[i] is the operation performed by the user at the ith step. ...

![https://leetcode.com/problems/crawler-log-folder/] The Leetcode file system keeps a log each time some user performs a change folder operation. The operations are described below: “../” : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder). “./” : Remain in the same folder. “x/” : Move to the child folder named x (This folder is guaranteed to always exist). You are given a list of strings logs where logs[i] is the operation performed by the user at the ith step. ...

![https://leetcode.com/problems/remove-outermost-parentheses/] A valid parentheses string is either empty (""), “(” + A + “)”, or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, “”, “()”, “(())()”, and “(()(()))” are all valid parentheses strings. A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings. ...

Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them. We repeatedly make duplicate removals on S until we no longer can. Return the final string after all such duplicate removals have been made. It is guaranteed the answer is unique. 1 2 3 4 5 6 Example 1: Input: "abbaca" Output: "ca" Explanation: For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move. The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca". Note: ...

![https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/] Given a string s, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them causing the left and the right side of the deleted substring to concatenate together. We repeatedly make k duplicate removals on s until we no longer can. Return the final string after all such duplicate removals have been made. It is guaranteed that the answer is unique. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Example 1: Input: s = "abcd", k = 2 Output: "abcd" Explanation: There's nothing to delete. Example 2: Input: s = "deeedbbcccbdaa", k = 3 Output: "aa" Explanation: First delete "eee" and "ccc", get "ddbbbdaa" Then delete "bbb", get "dddaa" Finally delete "ddd", get "aa" Example 3: Input: s = "pbbcggttciiippooaais", k = 2 Output: "ps" Constraints: ...

![https://leetcode.com/problems/min-stack/] Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. push(x) – Push element x onto stack. pop() – Removes the element on top of the stack. top() – Get the top element. getMin() – Retrieve the minimum element in the stack. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Example 1: Input ["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]] Output [null,null,null,null,-3,null,0,-2] Explanation MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); // return -3 minStack.pop(); minStack.top(); // return 0 minStack.getMin(); // return -2 Constraints: ...

![https://leetcode.com/problems/basic-calculator-ii/] Given a string s which represents an expression, evaluate this expression and return its value. The integer division should truncate toward zero. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Example 1: Input: s = "3+2*2" Output: 7 Example 2: Input: s = " 3/2 " Output: 1 Example 3: Input: s = " 3+5 / 2 " Output: 5 Constraints: ...

![https://leetcode.com/problems/backspace-string-compare/] Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character. Note that after backspacing an empty text, the text will continue empty. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Example 1: Input: S = "ab#c", T = "ad#c" Output: true Explanation: Both S and T become "ac". Example 2: Input: S = "ab##", T = "c#d#" Output: true Explanation: Both S and T become "". Example 3: Input: S = "a##c", T = "#a#c" Output: true Explanation: Both S and T become "c". Example 4: Input: S = "a#c", T = "b" Output: false Explanation: S becomes "c" while T becomes "b". Note: ...

ou are keeping score for a baseball game with strange rules. The game consists of several rounds, where the scores of past rounds may affect future rounds’ scores. At the beginning of the game, you start with an empty record. You are given a list of strings ops, where ops[i] is the ith operation you must apply to the record and is one of the following: An integer x - Record a new score of x. “+” - Record a new score that is the sum of the previous two scores. It is guaranteed there will always be two previous scores. “D” - Record a new score that is double the previous score. It is guaranteed there will always be a previous score. “C” - Invalidate the previous score, removing it from the record. It is guaranteed there will always be a previous score. Return the sum of all the scores on the record. ...