需求：實現一個簡單的計算器來分析一個簡單的表示式字串。

表示式字串可能包含括號,+ +或減號,非負整數和空格。

例子:“1 + 1 = 2,(1)“= 1(1 -(4 - 5)“= 2

問題分析：這個問題可以通過使用堆疊來解決。

網路配圖

Java解決方法

程式碼如下：

public int calculate(String s) {
// delte white spaces
s = s.replaceAll(" ", "");
Stack<String> stack = new Stack<String>();
char[] arr = s.toCharArray();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
if (arr[i] == ' ')
continue;
if (arr[i] >= '0' && arr[i] <= '9') {
sb.append(arr[i]);
if (i == arr.length - 1) {
stack.push(sb.toString());
}
} else {
if (sb.length() > 0) {
stack.push(sb.toString());
sb = new StringBuilder();
}
if (arr[i] != ')') {
stack.push(new String(new char[] { arr[i] }));
} else {
// when meet ')', pop and calculate
ArrayList<String> t = new ArrayList<String>();
while (!stack.isEmpty()) {
String top = stack.pop();
if (top.equals("(")) {
break;
} else {
t.add(0, top);
}
}
int temp = 0;
if (t.size() == 1) {
temp = Integer.valueOf(t.get(0));
} else {
for (int j = t.size() - 1; j > 0; j = j
- 2) {
if (t.get(j - 1).equals("-")) {
temp += 0 -
Integer.valueOf(t.get(j));
} else {
temp += Integer.valueOf
(t.get(j));
}
}
temp += Integer.valueOf(t.get(0));
}
stack.push(String.valueOf(temp));
}
}
}
ArrayList<String> t = new ArrayList<String>();
while (!stack.isEmpty()) {
String elem = stack.pop();
t.add(0, elem);
}
int temp = 0;
for (int i = t.size() - 1; i > 0; i = i - 2) {
if (t.get(i - 1).equals("-")) {
temp += 0 - Integer.valueOf(t.get(i));
} else {
temp += Integer.valueOf(t.get(i));
}
}
temp += Integer.valueOf(t.get(0));
return temp;
}