227. Basic Calculator II
Basic Calculator
Given a string expression s, please implement a basic calculator to calculate and return its value.
Integer division keeps only the integer part.
You can assume that a given expression is always valid. All intermediate results will be in the range [-231, 231 - 1].
Note: Any built-in functions that evaluate strings as mathematical expressions, such as eval(), are not allowed.
Example 1:
Input: s = "3+2*2"
Output: 7
the stack
First of all, the string must be disassembled, and the number and addition, subtraction, multiplication and division must be judged. The number needs to be converted from a character to a number. Then, when addition is encountered, the current num is added to the stack. Subtraction is the opposite number of the current number. , both multiplication and division need to take out the top element of the stack and then process it. Finally, the stack sum needs to be calculated.
step:
- iterate over the string
- judge numbers, convert
- Addition, add the number num, push to the stack
- Subtraction, add the opposite number of numbers, push to the stack
- Multiplication, the top element of the stack is multiplied by the number, and the stack is pushed
- Division, the top element of the stack is divided by the number, and the stack is pushed
- 栈和和
public int calculate(String s) {
Stack<Integer> stack = new Stack<>();
// 当前num值
int num = 0;
// 默认选择
char preSign = '+';
for(int i=0;i<s.length();i++){
// 判断字符是否是数字
if(Character.isDigit(s.charAt(i))){
// 转换为整数，多位字符进行拼接
num = num * 10 +s.charAt(i)-'0';
}
// 不是数字，空格，不是最后一个字符
if(!Character.isDigit(s.charAt(i)) && s.charAt(i) !=' ' || i==s.length()-1){
switch(preSign){
case '+':
// 添加
stack.push(num);
break;
case '-':
// 添加负数
stack.push(-num);
break;
case '*':
// 乘号去除栈顶元素和当前num相乘
stack.push(stack.pop()*num);
break;
case '/':
// 栈顶元素和当前num相除
stack.push(stack.pop()/num);
}
// 获取当前的字符，更新
preSign = s.charAt(i);
// num归0,重置
num=0;
}
}
// 计算之和
int ans = 0;
while(!stack.isEmpty()){
ans += stack.pop();
}
return ans;
}
Time complexity: O(n) traverses the string once, and then judge
Space complexity: O(n) The elements in the stack are taken out and added
Summary: This question did not consider the multi-digit situation at the beginning, and the method of judging whether the corresponding character is a number does not know how to judge.