題目描述：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
.
You may assume that the given expression is always valid.
Some examples:
"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.
解題思路：採用一個棧存儲，使用邊掃描字符串邊入棧並計算的過程，當輸入元素符合計算條件時，出棧並計算，計算後結果入棧。
Java代碼實現：
public class Solution {
public int calculate(String s) {
Stack<String> stack=new Stack<String>();
for(int i=0;i<s.length();i++){
char ch=s.charAt(i);
String strNum="";
if(ch!=' '){
if(ch=='+'||ch=='-'||ch=='('){
stack.push(String.valueOf(ch));
}
else if(ch>='0'&&ch<='9'){//保證多位數字
strNum=String.valueOf(ch);
while(i!=s.length()-1&&s.charAt(i+1)>='0'&&s.charAt(i+1)<='9'){
strNum+=s.charAt(++i);
}
while(!stack.isEmpty()){
String str=stack.peek();
if("(".equals(str)){
stack.push(strNum);
break;
}
if("+".equals(str)||"-".equals(str)){
String op=stack.pop();
if("+".equals(op)){
strNum=String.valueOf(Integer.valueOf(strNum)+Integer.valueOf(stack.pop()));
}
else if("-".equals(op)){
strNum=String.valueOf(Integer.valueOf(stack.pop())-Integer.valueOf(strNum));
}
}
}
if(stack.isEmpty()) stack.push(strNum);
}
else if(ch==')'){
strNum=stack.pop();
stack.pop();
while(!stack.isEmpty()){
String str=stack.peek();
if("(".equals(str)){
stack.push(strNum);
break;
}
if("+".equals(str)||"-".equals(str)){
String op=stack.pop();
if("+".equals(op)){
strNum=String.valueOf(Integer.valueOf(strNum)+Integer.valueOf(stack.pop()));
}
else if("-".equals(op)){
strNum=String.valueOf(Integer.valueOf(stack.pop())-Integer.valueOf(strNum));
}
}
}
if(stack.isEmpty()) stack.push(strNum);
}
}
}
return Integer.valueOf(stack.pop());
}
}版權聲明：本文爲博主原創文章，未經博主允許不得轉載。
