Question
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+22" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5
Code
public class Solution {
public int calculate(String s) {
if (s == null || s.length() == 0) return 0;
Stack<Integer> nums = new Stack<>();
Stack<Character> ops = new Stack<>();
StringBuilder sb = new StringBuilder();
int i = 0;
while (i < s.length()) {
char c = s.charAt(i);
if (c == ' ') {
i++;
continue;
}
if (c >= '0' && c <= '9') {
sb.append(c);
i++;
for (; i < s.length(); i++) {
char ci = s.charAt(i);
if (ci >= '0' && ci <= '9') {
sb.append(ci);
} else {
break;
}
}
nums.push(Integer.parseInt(sb.toString()));
sb.delete(0, sb.length());
continue;
}
if (c == '*' || c == '/') {
if (ops.isEmpty() || (ops.peek() == '+' || ops.peek() == '-')) {
ops.push(c);
}
else {
while (!ops.isEmpty() && ops.peek() != '+' && ops.peek() != '-') {
cal(nums, ops);
}
ops.push(c);
}
i++;
continue;
}
if (c == '+' || c == '-') {
while (!ops.isEmpty()) {
cal(nums, ops);
}
ops.push(c);
i++;
continue;
}
}
while (!ops.isEmpty()) {
cal(nums, ops);
}
return nums.pop();
}
public void cal(Stack<Integer> nums, Stack<Character> ops) {
char op = ops.pop();
int b = nums.pop();
int a = nums.pop();
nums.push(calculate(a, b, op));
}
public int calculate(int a, int b, char op) {
switch (op) {
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
return a / b;
}
return 0;
}
}
Solution
与Basic Calculator思路相似,利用数字栈和符号栈进行计算。
当符号为+或者-时,依次弹出符号栈中的符号并进行计算。当符号为或者/时,先判断栈顶符号,若栈顶符号为或者/则依次弹出符号栈中的符号进行计算,若栈顶符号为+或者-则直接入栈。
网友评论