Basic Calculator I && II && III

概述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 int

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 }

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