Basic Calculator I && II &&

发布时间：2020-12-14 21:50:32 所属栏目：大数据来源：网络整理

导读：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

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,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?),?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,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 }

（编辑：李大同）

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Basic Calculator I &amp;&amp; II &amp;&amp;

Basic Calculator I && II &&