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。
class Solution {
public int calculate(String s) {
int res = , num = , sign = , n = s.length();
for (int i = ; i < n; ++i) {
char c = s.charAt(i);
if (c >= '' && c <= '') {
num = * num + (c - '');
} else if (c == '(') {
int j = i, cnt = ;
for (; i < n; ++i) {
char letter = s.charAt(i);
if (letter == '(') ++cnt;
if (letter == ')') --cnt;
if (cnt == ) break;
}
num = calculate(s.substring(j + , i));
}
if (c == '+' || c == '-' || i == n - ) {
res += sign * num;
num = ;
sign = (c == '+') ? : -;
}
}
return res;
}
}
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。
注意：我们处理的时候，总是处理之前一个符号，而不是当前符号
class Solution {
public int calculate(String s) {
int res = , num = , n = s.length();
char op = '+';
Stack<Integer> st = new Stack<>();
for (int i = ; i < n; ++i) {
char ch = s.charAt(i);
if (Character.isDigit(ch)) {
num = num * + ch - '';
} else if (isOperator(ch)) {
addToStack(op, num, st);
op = ch;
num = ;
}
}
// handle last case
addToStack(op, num, st);
while (!st.empty()) {
res += st.pop();
}
return res;
}
private boolean isOperator(char ch) {
if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {
return true;
}
return false;
}
private void addToStack(char op, int num, Stack<Integer> st) {
if (op == '+') st.push(num);
if (op == '-') st.push(-num);
if (op == '*' || op == '/') {
int tmp = (op == '*') ? st.pop() * num : st.pop() / num;
st.push(tmp);
}
}
}
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。
class Solution {
public int calculate(String s) {
int res = , num = , n = s.length();
char op = '+';
Stack<Integer> st = new Stack<>();
for (int i = ; i < n; ++i) {
char ch = s.charAt(i);
if (Character.isDigit(ch)) {
num = num * + ch - '';
} else if (ch == '(') {
int j = i, cnt = ;
for (; i < n; ++i) {
char letter = s.charAt(i);
if (letter == '(') ++cnt;
if (letter == ')') --cnt;
if (cnt == ) break;
}
num = calculate(s.substring(j + , i));
} else if (isOperator(ch)) {
addToStack(op, num, st);
op = ch;
num = ;
}
}
// handle last case
addToStack(op, num, st);
while (!st.empty()) {
res += st.pop();
}
return res;
}
private static boolean isOperator(char ch) {
if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {
return true;
}
return false;
}
private static void addToStack(char op, int num, Stack<Integer> st) {
if (op == '+') st.push(num);
if (op == '-') st.push(-num);
if (op == '*' || op == '/') {
int tmp = (op == '*') ? st.pop() * num : st.pop() / num;
st.push(tmp);
}
}
}
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