LeetCode——基本计算器

题目描述:

给你一个字符串表达式 s ,请你实现一个基本计算器来计算并返回它的值。

提示:

  • 1 <= s.length <= 3 * 105
  • s 由数字、’+’、’-’、’(’、’)’、和 ’ ’ 组成
  • s 表示一个有效的表达式

示例 1:
输入:s = “1 + 1”
输出:2

示例 2:
输入:s = " 2-1 + 2 "
输出:3

示例 3:
输入:s = “(1+(4+5+2)-3)+(6+8)”
输出:23

代码如下:

class Solution {
    
    
    public int calculate(String s) {
    
    
        Deque<Character> deque = new LinkedList<>();
        for (char c : s.toCharArray()) {
    
    
            deque.offer(c);
        }
        return sum(deque);
    }

    public int sum(Deque<Character> deque) {
    
    
        Stack<Integer> stack = new Stack<>();
        char sign = '+';
        int num = 0;
        int res = 0;
        while (!deque.isEmpty()) {
    
    
            char c = deque.poll();
            if (Character.isDigit(c)) {
    
    
                num = num * 10 + c - '0';
            }
            if (c == '(') {
    
    
                num = sum(deque);
            }
            if ((!Character.isDigit(c) && c != ' ') || deque.isEmpty()) {
    
    
                if (sign == '+') {
    
    
                    stack.push(num);
                } else if (sign == '-') {
    
    
                    stack.push(-num);
                } else if (sign == '*') {
    
    
                    stack.push(stack.pop() * num);
                } else if (sign == '/') {
    
    
                    stack.push(stack.pop() / num);
                }
                num = 0;
                sign = c;
            }
            if (c == ')') {
    
    
                break;
            }
        }
        for (int i : stack) {
    
    
            res += i;
        }
        return res;
    }
}

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转载自blog.csdn.net/FYPPPP/article/details/114655419