[LeetCode] 227. Basic Calculator II 基本计算器 II
Implement a basic calculator to evaluate a simple expression string.
The expression string contains only non-negative integers, +
, -
, *
, /
operators and empty spaces . The integer division should truncate toward zero.
You may assume that the given expression is always valid.
Some examples:
"3+2*2" = 7 " 3/2 " = 1 " 3+5 / 2 " = 5
Note: Do not use the eval
built-in library function.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
Java:
public int calculate(String s) { int md=-1; // 0 is m, 1 is d int sign=1; // 1 is +, -1 is - int prev=0; int result=0; for(int i=0; i<s.length(); i++){ char c = s.charAt(i); if(Character.isDigit(c)){ int num = c-'0'; while(++i<s.length() && Character.isDigit(s.charAt(i))){ num = num*10+s.charAt(i)-'0'; } i--; // back to last digit of number if(md==0){ prev = prev * num; md=-1; }else if(md==1){ prev = prev / num; md=-1; }else{ prev = num; } }else if(c=='/'){ md=1; }else if(c=='*'){ md=0; }else if(c=='+'){ result = result + sign*prev; sign=1; }else if(c=='-'){ result = result + sign*prev; sign=-1; } } result = result + sign*prev; return result; }
Python:
class Solution: # @param {string} s # @return {integer} def calculate(self, s): operands, operators = [], [] operand = "" for i in reversed(xrange(len(s))): if s[i].isdigit(): operand += s[i] if i == 0 or not s[i-1].isdigit(): operands.append(int(operand[::-1])) operand = "" elif s[i] == ')' or s[i] == '*' or s[i] == '/': operators.append(s[i]) elif s[i] == '+' or s[i] == '-': while operators and \ (operators[-1] == '*' or operators[-1] == '/'): self.compute(operands, operators) operators.append(s[i]) elif s[i] == '(': while operators[-1] != ')': self.compute(operands, operators) operators.pop() while operators: self.compute(operands, operators) return operands[-1] def compute(self, operands, operators): left, right = operands.pop(), operands.pop() op = operators.pop() if op == '+': operands.append(left + right) elif op == '-': operands.append(left - right) elif op == '*': operands.append(left * right) elif op == '/': operands.append(left / right)
C++:
class Solution { public: int calculate(string s) { int res = 0, d = 0; char sign = '+'; stack<int> nums; for (int i = 0; i < s.size(); ++i) { if (s[i] >= '0') { d = d * 10 + s[i] - '0'; } if ((s[i] < '0' && s[i] != ' ') || i == s.size() - 1) { if (sign == '+') nums.push(d); if (sign == '-') nums.push(-d); if (sign == '*' || sign == '/') { int tmp = sign == '*' ? nums.top() * d : nums.top() / d; nums.pop(); nums.push(tmp); } sign = s[i]; d = 0; } } while (!nums.empty()) { res += nums.top(); nums.pop(); } return res; } };
C++:
class Solution { public: int calculate(string s) { stack<int64_t> operands; stack<char> operators; string operand; for (int i = s.length() - 1; i >= 0; --i) { if (isdigit(s[i])) { operand.push_back(s[i]); if (i == 0 || !isdigit(s[i - 1])) { reverse(operand.begin(), operand.end()); operands.emplace(stol(operand)); operand.clear(); } } else if (s[i] == ')' || s[i] == '*' || s[i] == '/') { operators.emplace(s[i]); } else if (s[i] == '+' || s[i] == '-') { while (!operators.empty() && (operators.top() == '*' || operators.top() == '/')) { compute(operands, operators); } operators.emplace(s[i]); } else if (s[i] == '(') { // operators at least one element, i.e. ')'. while (operators.top() != ')') { compute(operands, operators); } operators.pop(); } } while (!operators.empty()) { compute(operands, operators); } return operands.top(); } void compute(stack<int64_t>& operands, stack<char>& operators) { const int64_t left = operands.top(); operands.pop(); const int64_t right = operands.top(); operands.pop(); const char op = operators.top(); operators.pop(); if (op == '+') { operands.emplace(left + right); } else if (op == '-') { operands.emplace(left - right); } else if (op == '*') { operands.emplace(left * right); } else if (op == '/') { operands.emplace(left / right); } } };
类似题目:
[LeetCode] 224. Basic Calculator 基本计算器
All LeetCode Questions List 题目汇总