花花酱 LeetCode 227. Basic Calculator II

zxi — Tue, 10 Mar 2020 01:24:59 +0000

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.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

You may assume that the given expression is always valid.
Do not use the eval built-in library function.

Solution: Stack

if operator is ‘+’ or ‘-’, push the current num * sign onto stack.
if operator ‘*’ or ‘/’, pop the last num from stack and * or / by the current num and push it back to stack.

The answer is the sum of numbers on stack.

3+2*2 => {3}, {3,2}, {3, 2*2} = {3, 4} => ans = 7
3 +5/2 => {3}, {3,5}, {3, 5/2} = {3, 2} => ans = 5
1 + 2*3 – 5 => {1}, {1,2}, {1,2*3} = {1,6}, {1, 6, -5} => ans = 2

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua
class Solution {
public:
  int calculate(string s) {
    vector nums;    
    char op = '+';
    int cur = 0;
    int pos = 0;
    while (pos < s.size()) {
      if (s[pos] == ' ') {
        ++pos;
        continue;
      }
      while (isdigit(s[pos]) && pos < s.size())
        cur = cur * 10 + (s[pos++] - '0');      
      if (op == '+' || op == '-') {
        nums.push_back(cur * (op == '+' ? 1 : -1));
      } else if (op == '*') {
        nums.back() *= cur;
      } else if (op == '/') {
        nums.back() /= cur;
      }
      cur = 0;      
      op = s[pos++];
    }
    return accumulate(begin(nums), end(nums), 0);
  }
};

python3

# Author: Huahua
class Solution:
  def calculate(self, s: str) -> int:
    nums = []
    op = '+'
    cur = 0
    i = 0
    while i < len(s):
      if s[i] == ' ': 
        i += 1
        continue
      while i < len(s) and s[i].isdigit():
        cur = cur * 10 + ord(s[i]) - ord('0')
        i += 1      
      if op in '+-':
        nums.append(cur * (1 if op == '+' else -1))
      elif op == '*':
        nums[-1] *= cur
      elif op == '/':
        sign = -1 if nums[-1] < 0 or cur < 0 else 1
        nums[-1] = abs(nums[-1]) // abs(cur) * sign
      cur = 0
      if (i < len(s)): op = s[i]
      i += 1    
    return sum(nums)

花花酱 LeetCode 224. Basic Calculator

zxi — Mon, 09 Mar 2020 20:33:55 +0000

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.

Solution: Recursion

Make a recursive call when there is an open parenthesis and return if there is close parenthesis.

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua
class Solution {
public:
  int calculate(string s) {    
    function eval = [&](int& pos) {
      int ret = 0;
      int sign = 1;
      int val = 0;
      while (pos < s.size()) {
        const char ch = s[pos];
        if (isdigit(ch)) {
          val = val * 10 + (s[pos++] - '0');
        } else if (ch == '+' || ch == '-') {
          ret += sign * val;
          val = 0;
          sign = ch == '+' ? 1 : -1;
          ++pos;
        } else if (ch == '(') {
          val = eval(++pos);
        } else if (ch == ')') {          
          ++pos;
          break;
        } else {
          ++pos;
        }
      }
      return ret += sign * val;
    };
    int pos = 0;
    return eval(pos);
  }
};

Python3

# Author: Huahua
class Solution:
  def calculate(self, s: str) -> int:
    self.pos = 0    
    def eval():
      val = 0
      sign = 1
      ret = 0
      while self.pos < len(s):
        ch = s[self.pos]
        if ch == ' ':
          self.pos += 1
        elif ch == '(':
          self.pos += 1
          val = eval()
        elif ch == ')':
          self.pos += 1
          ret += sign * val
          return ret
        elif ch == '+' or ch == '-':
          ret += sign * val
          val = 0
          sign = 1 if ch == '+' else -1
          self.pos += 1
        else:
          val = val * 10 + (ord(ch) - ord('0'))
          self.pos += 1
      ret += val * sign
      return ret
    return eval()

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