Basic Calculator I && II && III

Basic Calculator I

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.

分析：这题因为不存在乘法和除法，所以，对于里面的减号，我们可以把它当成+（-num）来处理。所以，每次遇到一个数字的时候，我们需要知道这个数字的符号，每当我们把这个数字所有的digit都拿到以后，就可以得到这个数，然后把这个数加到之前的临时结果里。

对于比较特殊的处理是括号，但是这里有一个很巧的思路，我们可以把括号里的表达式call当前的方法来计算。

 1 class Solution {
 2     public int calculate(String s) {
 3         int res = 0, num = 0, sign = 1, n = s.length();
 4         for (int i = 0; i < n; ++i) {
 5             char c = s.charAt(i);
 6             if (c >= ‘0‘ && c <= ‘9‘) {
 7                 num = 10 * num + (c - ‘0‘);
 8             } else if (c == ‘(‘) {
 9                 int j = i, cnt = 0;
10                 for (; i < n; ++i) {
11                     char letter = s.charAt(i);
12                     if (letter == ‘(‘) ++cnt;
13                     if (letter == ‘)‘) --cnt;
14                     if (cnt == 0) break;
15                 }
16                 num = calculate(s.substring(j + 1, i));
17             }
18             if (c == ‘+‘ || c == ‘-‘ || i == n - 1) {
19                 res += sign * num;
20                 num = 0;
21                 sign = (c == ‘+‘) ? 1 : -1;
22              } 
23         }
24         return res;
25     }
26 }

 Basic Calculator II

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.

分析：因为没有括号，所以我们可以把加或者减的部分当成一个数，比如 5-2，把它当成（5）+（-2）。同理，对于有乘或者除，或者既有乘又有除的话，也把它当成一个数，比如5-3*2/4=（5）-（3*2/4）。对于乘法和除法，我们总是从左算到右，所以我们可以把* or /之前的部分先存下来，当符号是 * or /的时候，再取出来就可以了。

 1 class Solution {
 2     public int calculate(String s) {
 3         int res = 0, num = 0, n = s.length();
 4         char op = ‘+‘;
 5         Stack<Integer> st = new Stack<>();
 6         for (int i = 0; i < n; ++i) {
 7             char ch = s.charAt(i);
 8             if (Character.isDigit(ch)) {
 9                 num = num * 10 + ch - ‘0‘;
10             }
11             if ((ch < ‘0‘ && ch != ‘ ‘) || i == n - 1) {
12                 if (op == ‘+‘) st.push(num);
13                 if (op == ‘-‘) st.push(-num);
14                 if (op == ‘*‘ || op == ‘/‘) {
15                     int tmp = (op == ‘*‘) ? st.pop() * num : st.pop() / num;
16                     st.push(tmp);
17                 }
18                 op = ch;
19                 num = 0;
20             } 
21         }
22         while (!st.empty()) {
23             res += st.pop();
24         }
25         return res;
26     }
27 }

 Basic Calculator III

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-negativeintegers and empty spaces .

The expression string contains only non-negative integers, +, -, *, / operators , open ( and closing parentheses ) and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-2147483648, 2147483647].

Some examples:

"1 + 1" = 2
" 6-4 / 2 " = 4
"2*(5+5*2)/3+(6/2+8)" = 21
"(2+6* 3+5- (3*14/7+2)*5)+3"=-12

Note: Do not use the eval built-in library function.

分析：只要把括号部分的处理加进来就可以了。

 1 class Solution {
 2     public static int calculate(String s) {
 3         int res = 0, num = 0, n = s.length();
 4         char op = ‘+‘;
 5         Stack<Integer> st = new Stack<>();
 6         for (int i = 0; i < n; ++i) {
 7             char ch = s.charAt(i);
 8             if (Character.isDigit(ch)) {
 9                 num = num * 10 + ch - ‘0‘;
10             } else if (ch == ‘(‘) {
11                 int j = i, cnt = 0;
12                 for (; i < n; ++i) {
13                     char letter = s.charAt(i);
14                     if (letter == ‘(‘) ++cnt;
15                     if (letter == ‘)‘) --cnt;
16                     if (cnt == 0) break;
17                 }
18                 num = calculate(s.substring(j + 1, i));
19             }
20             if ((ch < ‘0‘ && ch != ‘ ‘) || i == n - 1) {
21                 if (op == ‘+‘) st.push(num);
22                 if (op == ‘-‘) st.push(-num);
23                 if (op == ‘*‘ || op == ‘/‘) {
24                     int tmp = (op == ‘*‘) ? st.pop() * num : st.pop() / num;
25                     st.push(tmp);
26                 }
27                 op = ch;
28                 num = 0;
29             }
30         }
31         while (!st.empty()) {
32             res += st.pop();
33         }
34         return res;
35     }
36 }