标签：
Description：
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.
用两个操作栈来处理，具体见代码。
public class Solution {
public static int calculate(String s) {
Stack<Integer> num = new Stack<>();
Stack<Character> op = new Stack<>();
op.push(‘+‘);
int n = s.length();
if(n < 1) return 0;
for(int i=0; i<n; ) {
if(s.charAt(i) == ‘ ‘) {
i ++;
}
else if(s.charAt(i) == ‘(‘) {
op.push(‘(‘);
op.push(‘+‘);
i ++;
}
else if(isOp(s.charAt(i))) {
op.push(s.charAt(i));
i ++;
}
//523 +-(++
else if(s.charAt(i) == ‘)‘) {
int res = 0;
while (!op.empty() && !(op.peek() == ‘(‘)) {
int temp = num.peek();
if (op.peek() == ‘-‘) temp = -temp;
res += temp;
num.pop();
op.pop();
}
System.out.println(res);
op.pop();
num.push(res);
i ++;
}
else {
int temp = 0;
while(i < n && isDigit(s.charAt(i))) {
temp = temp * 10;
temp += s.charAt(i) - ‘0‘;
i ++;
}
num.push(temp);
}
}
int res = 0;
while (!op.empty()) {
int temp = num.peek();
if (op.peek() == ‘-‘) temp = -temp;
res += temp;
num.pop();
op.pop();
}
return res;
}
public static boolean isOp(char c) {
if(c == ‘+‘ || c == ‘-‘) {
return true;
}
else {
return false;
}
}
public static boolean isDigit(char c) {
if(c >= ‘0‘ && c <=‘9‘) {
return true;
}
else {
return false;
}
}
}
标签：
原文地址：http://www.cnblogs.com/wxisme/p/4614848.html
