标签:min else tput sig ati div problems git exp
package LeetCode_224 import java.util.* /** * 224. Basic Calculator * https://leetcode.com/problems/basic-calculator/description/ * * Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . Example 1: Input: "1 + 1" Output: 2 Example 2: Input: " 2-1 + 2 " Output: 3 Example 3: Input: "(1+(4+5+2)-3)+(6+8)" Output: 23 Note: You may assume that the given expression is always valid. Do not use the eval built-in library function. * */ class Solution { private val operator = listOf("+", "-") fun calculate(s: String): Int { val stack = Stack<String>() var i = s.length - 1 while (i >= 0) { if (s[i]==‘ ‘){ i-- continue } if (s[i] == ‘)‘ || s[i] == ‘+‘ || s[i] == ‘-‘) { stack.push(s[i].toString()) } else if (s[i] == ‘(‘) { val sum = getSum(stack) stack.pop() stack.push(sum.toString()) } else { val intSB = StringBuilder() //pick out the digit, for example:10 while (i >= 0 && s[i].isDigit()) { intSB.insert(0,s[i--]) } i++//prevent lose some operator, so need go back, for example: 2-1 + 2 stack.push(intSB.toString()) } i-- } //result = getSum(stack) return getSum(stack) } private fun getSum(stack: Stack<String>): Int { var sign = "+" var total = 0 while (stack.isNotEmpty() && stack.peek() != "(" && stack.peek() != ")") { if (stack.peek() in operator) { sign = stack.pop() } else { val temp = stack.pop().toInt() total += if (sign == "+") temp else -temp } } return total } }
标签:min else tput sig ati div problems git exp
原文地址:https://www.cnblogs.com/johnnyzhao/p/13121345.html