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Problem solution · Java

Basic Calculator III

Basic Calculator III: a Java solution using stack-based processing. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Stack-based processing
Source
walkccc LeetCode Solutions
Length
59 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Stack-based processing

For Basic Calculator III, the implementation keeps unresolved items in last-in, first-out order, often to match boundaries, parse structure, or maintain monotonic candidates.

  1. Define what every stack entry represents.
  2. Pop entries once the current item resolves or invalidates them.
  3. Push the current item with only the information later steps need.

Code notes

  • 59 lines of Java from the credited upstream file 772.java.
  • The implementation visibly relies on work queue.
  • 5 loop blocks detected.

Complexity

If each item is pushed and popped at most once, the stack work is linear.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeBasic Calculator III · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int calculate(String s) {    Deque<Integer> nums = new ArrayDeque<>();    Deque<Character> ops = new ArrayDeque<>();    boolean hasPrevNum = false;     for (int i = 0; i < s.length(); ++i) {      final char c = s.charAt(i);      if (Character.isDigit(c)) {        int num = c - '0';        while (i + 1 < s.length() && Character.isDigit(s.charAt(i + 1)))          num = num * 10 + (s.charAt(i++ + 1) - '0');        nums.push(num);        hasPrevNum = true;      } else if (c == '(') {        ops.push('(');        hasPrevNum = false;      } else if (c == ')') {        while (ops.peek() != '(')          calc(nums, ops);        ops.pop(); // Pop '('.      } else if (c == '+' || c == '-' || c == '*' || c == '/') {        if (!hasPrevNum)          nums.push(0);        while (!ops.isEmpty() && precedes(ops.peek(), c))          calc(nums, ops);        ops.push(c);      }    }     while (!ops.isEmpty())      calc(nums, ops);     return nums.peek();  }   private void calc(Deque<Integer> nums, Deque<Character> ops) {    final int b = nums.pop();    final int a = nums.pop();    final char op = ops.pop();    if (op == '+')      nums.push(a + b);    else if (op == '-')      nums.push(a - b);    else if (op == '*')      nums.push(a * b);    else // op == '/'      nums.push(a / b);  }   // Returns true if the previous character is a operator and the priority of  // the previous operator >= the priority of the current character (operator).  private boolean precedes(char prev, char curr) {    if (prev == '(')      return false;    return prev == '*' || prev == '/' || curr == '+' || curr == '-';  }}

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