标签：
Description：
Implement a basic calculator to evaluate a simple expression string.
The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.
You may assume that the given expression is always valid.
Some examples:
"3+2*2" = 7 " 3/2 " = 1 " 3+5 / 2 " = 5
Note: Do not use the eval built-in library function.
用两个操作栈来处理数据和操作符。先计算*和/然后变成加减法后再按顺序进行计算。
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(isOp(s.charAt(i))) {
op.push(s.charAt(i));
i ++;
}
else {
int temp = 0;
while(i < n && isDigit(s.charAt(i))) {
temp = temp * 10;
temp += s.charAt(i) - ‘0‘;
i ++;
}
if(!op.empty() && (op.peek()==‘*‘ || op.peek()==‘/‘)) {
if(op.peek() == ‘*‘) {
temp *= num.pop();
op.pop();
}
else {
temp = num.pop() / temp;
op.pop();
}
}
num.push(temp);
}
}
int res = 0;
while (!op.empty()) {
int temp = num.peek();
//System.out.println(temp);
if (op.peek() == ‘-‘) temp = -temp;
res += temp;
num.pop();
op.pop();
}
return res;
}
public static boolean isOp(char c) {
if(c == ‘+‘ || c == ‘-‘ || 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/4614857.html
