Algorithm
Problem Name: 227. Basic Calculator II
Given a string s which represents an expression, evaluate this expression and return its value.
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 [-231, 231 - 1].
Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().
Example 1:
Input: s = "3+2*2" Output: 7
Example 2:
Input: s = " 3/2 " Output: 1
Example 3:
Input: s = " 3+5 / 2 " Output: 5
Constraints:
1 <= s.length <= 3 * 105sconsists of integers and operators('+', '-', '*', '/')separated by some number of spaces.srepresents a valid expression.- All the integers in the expression are non-negative integers in the range
[0, 231 - 1]. - The answer is guaranteed to fit in a 32-bit integer.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
typedef struct {
int *p;
int sz;
int n;
} s_t;
int parse(char **sp, int *k) {
char *s = *sp;
while (*s == ' ') s ++;
*k = 0;
if (*s == 0) return 0;
if (*s == '+' || *s == '-' || *s == '*' || *s == '/') {
*k = *s == '+' ? 1 :
*s == '-' ? 2 :
*s == '*' ? 3 : 4;
*sp = ++ s;
return 1;
}
while (*s >= '0' && *s < = '9') {
*k = (*k) * 10 + *s - '0';
s ++;
}
*sp = s;
return 2;
}
void push(s_t *stack, int k) {
if (stack->sz == stack->n) {
stack->sz *= 2;
stack->p = realloc(stack->p, stack->sz * sizeof(int));
//assert(stack->p);
}
stack->p[stack->n ++] = k;
}
int low_op(s_t *ops, int k) {
const int priority[] = { 0, 1, 1, 2, 2 }; // null, +, -, *, /
return (priority[ops->p[ops->n - 1]] >= priority[k]) ? 1 : 0;
}
int calculate(char* s) {
s_t data = { 0 }, ops = { 0 };
int x, k, d1, d2, o;
data.n = ops.n = 0;
data.sz = ops.sz = 10;
data.p = malloc(data.sz * sizeof(int));
ops.p = malloc(ops.sz * sizeof(int));
push(&data, 0); // put a zero in case of with a null input
push(&ops, 0); // put a null operator on top of operator stack
do {
x = parse(&s, &k);
if (x == 2) { // data, push to stack
push(&data, k);
} else { // operator
while (low_op(&ops, k)) {
o = ops.p[-- ops.n];
if (o == 0) break; // the end
d2 = data.p[-- data.n];
d1 = data.p[-- data.n];
switch (o) {
case 1: // '+'
d1 = d1 + d2;
break;
case 2: // '-'
d1 = d1 - d2;
break;
case 3: // '*'
d1 = d1 * d2;
break;
case 4: // '/'
d1 = d1 / d2;
break;
default:
break;
}
push(&data, d1);
}
if (k) push(&ops, k);
}
} while (x);
//assert(ops.n == 0);
k = data.p[data.n - 1];
free(data.p);
free(ops.p);
return k;
}
Input
s = "3+2*2"
Output
7
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
private:
unordered_mapstk;
int tmp = 0;
char op = '+';
for(int i = 0; i < s.size(); i++){
char c = s[i];
if(c == ' ') continue;
if(isdigit(c)) tmp = tmp*10 + c - '0';
if(!isdigit(c)){
if(op == '+')
stk.push(tmp);
else if(op == '-')
stk.push(-tmp);
else if(op == '*'){
int n = stk.top();
stk.pop();
stk.push(n*tmp);
}
else if(op == '/'){
int n = stk.top();
stk.pop();
stk.push(n/tmp);
}
op = c;
tmp = 0;
}
}
int res = 0;
while(!stk.empty()) res += stk.top(), stk.pop();
return res;
}
};
Input
s = "3+2*2"
Output
7
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int calculate(String s) {
int currNum = 0;
int prevNum = 0;
int result = 0;
char operation = '+';
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
currNum = currNum * 10 + Character.getNumericValue(c);
}
if ((!Character.isDigit(c) && !Character.isWhitespace(c)) || i == s.length() - 1) {
if (operation == '-' || operation == '+') {
result += prevNum;
prevNum = operation == '+' ? currNum : -currNum;
} else if (operation == '*') {
prevNum = prevNum * currNum;
} else {
prevNum = prevNum / currNum;
}
operation = c;
currNum = 0;
}
}
result += prevNum;
return result;
}
}
Input
s = " 3/2 "
Output
1
#4 Code Example with Javascript Programming
Code -
Javascript Programming
start coding...Input
s = " 3/2 "
Output
1
#5 Code Example with Python Programming
Code -
Python Programming
const calculate = function(s) {
const stk = []
let op = '+', num = 0
s = s.trim()
const isDigit = ch => ch >= '0' && ch <= '9'
for(let i = 0, n = s.length; i < n; i++) {
const ch = s[i]
if(ch === ' ') continue
if(isDigit(ch)) {
num = (+num) * 10 + (+ch)
}
if(!isDigit(ch) || i === n - 1) {
if(op === '-') stk.push(-num)
else if(op === '+') stk.push(num)
else if(op === '*') stk.push(stk.pop() * num)
else if(op === '/') stk.push(~~(stk.pop() / num))
op = ch
num = 0
}
}
let res = 0
for(const e of stk) res += e
return res
};
Input
s = " 3+5 / 2 "
Output
5
#6 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0227_BasicCalculatorII
{
public int Calculate(string s)
{
char sign = '+';
var nums = new Stack < int>();
int cur = 0;
for (int i = 0; i < = s.Length; i++)
{
if (i < s.Length && s[i] >= '0' && s[i] <= '9')
cur = cur * 10 + s[i] - '0';
else if (i == s.Length || s[i] != ' ')
{
if (sign == '+') nums.Push(cur);
else if (sign == '-') nums.Push(-cur);
else if (sign == '*')
{
var pre = nums.Pop();
nums.Push(pre * cur);
}
else if (sign == '/')
{
var pre = nums.Pop();
nums.Push(pre / cur);
}
if (i < s.Length)
{
cur = 0;
sign = s[i];
}
}
}
var res = 0;
while (nums.Count > 0)
res += nums.Pop();
return res;
}
}
}
Input
s = " 3+5 / 2 "
Output
5