[LeetCode] 224. Basic Calculator 基本计算器

2021年09月15日 阅读数:1
这篇文章主要向大家介绍[LeetCode] 224. Basic Calculator 基本计算器,主要内容包括基础应用、实用技巧、原理机制等方面,希望对大家有所帮助。

Implement a basic calculator to evaluate a simple expression string.html

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -non-negative integers and empty spaces .java

You may assume that the given expression is always valid.python

Some examples:git

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

Note: Do not use the eval built-in library function.express

 

Java:app

public int calculate(String s) {
	// delte white spaces
	s = s.replaceAll(" ", "");
 
	Stack<String> stack = new Stack<String>();
	char[] arr = s.toCharArray();
 
	StringBuilder sb = new StringBuilder();
	for (int i = 0; i < arr.length; i++) {
		if (arr[i] == ' ')
			continue;
 
		if (arr[i] >= '0' && arr[i] <= '9') {
			sb.append(arr[i]);
 
			if (i == arr.length - 1) {
				stack.push(sb.toString());
			}
		} else {
			if (sb.length() > 0) {
				stack.push(sb.toString());
				sb = new StringBuilder();
			}
 
			if (arr[i] != ')') {
				stack.push(new String(new char[] { arr[i] }));
			} else {
				// when meet ')', pop and calculate
				ArrayList<String> t = new ArrayList<String>();
				while (!stack.isEmpty()) {
					String top = stack.pop();
					if (top.equals("(")) {
						break;
					} else {
						t.add(0, top);
					}
				}
 
				int temp = 0;
				if (t.size() == 1) {
					temp = Integer.valueOf(t.get(0));
				} else {
					for (int j = t.size() - 1; j > 0; j = j - 2) {
						if (t.get(j - 1).equals("-")) {
							temp += 0 - Integer.valueOf(t.get(j));
						} else {
							temp += Integer.valueOf(t.get(j));
						}
					}
					temp += Integer.valueOf(t.get(0));
				}
				stack.push(String.valueOf(temp));
			}
		}
	}
 
	ArrayList<String> t = new ArrayList<String>();
	while (!stack.isEmpty()) {
		String elem = stack.pop();
		t.add(0, elem);
	}
 
	int temp = 0;
	for (int i = t.size() - 1; i > 0; i = i - 2) {
		if (t.get(i - 1).equals("-")) {
			temp += 0 - Integer.valueOf(t.get(i));
		} else {
			temp += Integer.valueOf(t.get(i));
		}
	}
	temp += Integer.valueOf(t.get(0));
 
	return temp;
}  

Python:post

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] == '(':
                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)

C++:ui

class Solution {
public:
    int calculate(string s) {
        int res = 0, sign = 1, n = s.size();
        stack<int> st;
        for (int i = 0; i < n; ++i) {
            char c = s[i];
            if (c >= '0') {
                int num = 0;
                while (i < n && s[i] >= '0') {
                    num = 10 * num + s[i++] - '0';
                }
                res += sign * num;
                --i;
            } else if (c == '+') {
                sign = 1;
            } else if (c == '-') {
                sign = -1;
            } else if (c == '(') {
                st.push(res);
                st.push(sign);
                res = 0;
                sign = 1;
            } else if (c == ')') {
                res *= st.top(); st.pop();
                res += st.top(); st.pop();
            }
        }
        return res;
    }
};  

C++:lua

class Solution2 {
public:
    int calculate(string s) {
        stack<int> 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(stoi(operand));
                    operand.clear();
                }
            } else if (s[i] == ')' || s[i] == '+' || s[i] == '-') {
                operators.emplace(s[i]);
            } else if (s[i] == '(') {
                while (operators.top() != ')') {
                    compute(operands, operators);
                }
                operators.pop();
            }
        }
        while (!operators.empty()) {
            compute(operands, operators);
        }
        return operands.top();
    }

    void compute(stack<int>& operands, stack<char>& operators) {
        const int left = operands.top();
        operands.pop();
        const int 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);
        }
    } 
};

    

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