LeetCode 题解（113）: Basic Calculator II

题目：

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.

题解：

核心是找下一个操作数，操作数可是一个字符或多个字符。注意若为空格或“+”皆可忽略，最后计算堆栈中所有操作数的和。

class Solution {public:int calculate(string s) {queue<char> sign;stack<int> operand;for (int i = 0; i < s.length(); i++) {if (isdigit(s[i])) {getNextNumber(i, s, operand);}else if (s[i] == '-') {char current = s[i];i++;while (!isdigit(s[i]))i++;getNextNumber(i, s, operand);if (current == '-') {int old = operand.top();operand.pop();operand.push(old * (-1));}}else if (s[i] == ' ' || s[i] == '+') {continue;}else {int first = operand.top();operand.pop();int current = i;i++;while (!isdigit(s[i]))i++;getNextNumber(i, s, operand);int second = operand.top();operand.pop();if (s[current] == '*') {operand.push(first * second);}else {operand.push(first / second);}}}int result = operand.top();operand.pop();while (!operand.empty()) {result += operand.top();operand.pop();}return result;}void getNextNumber(int& pos, string& s, stack<int>& operand) {int result = s[pos++] - '0';while (isdigit(s[pos]) || s[pos] == ' ') {if (s[pos] == ' ') {pos++;continue;}else {result = result * 10 + (s[pos] - '0');pos++;}}operand.push(result);pos--;}};

Java版：

public class Solution {    public int calculate(String s) {        Stack<Integer> operand = new Stack<>();        for(int i = 0; i < s.length(); i++) {            if(Character.isDigit(s.charAt(i))) {                i = getNextNumber(s, i, operand);            } else if(s.charAt(i) == ' ' || s.charAt(i) == '+') {                continue;            } else if(s.charAt(i) == '-') {                i++;                while(!Character.isDigit(s.charAt(i)))                    i++;                i = getNextNumber(s, i, operand);                int old = operand.pop();                operand.push(old * (-1));            } else {                int first = operand.pop();                int current = i;                i++;                while(!Character.isDigit(s.charAt(i)))                    i++;                i = getNextNumber(s, i, operand);                int second = operand.pop();                if(s.charAt(current) == '*') {                    operand.push(first * second);                } else {                    operand.push(first / second);                }            }        }                            int result = operand.pop();        while(!operand.empty())            result += operand.pop();        return result;    }        public int getNextNumber(String s, int pos, Stack<Integer> operand) {        int result = s.charAt(pos++) - '0';        while(pos < s.length() && (Character.isDigit(s.charAt(pos)) || s.charAt(pos) == ' ')) {            if(s.charAt(pos) == ' ') {                pos++;                continue;            } else {                result = result * 10 + (s.charAt(pos) - '0');                pos++;            }        }        operand.push(result);        pos -= 1;        return pos;    }}

Python版：

class Solution:    # @param {string} s    # @return {integer}    def calculate(self, s):        operand = []        i = 0        while i < len(s):            if s[i].isdigit():                i = self.findNextNumber(s, i, operand)            elif s[i] == ' ' or s[i] == '+':                i += 1                continue            elif s[i] == '-':                i += 1                while not s[i].isdigit():                    i += 1                i = self.findNextNumber(s, i, operand)                old = operand[-1]                operand.pop()                operand.append(old * (-1))            else:                current = i                i += 1                first = operand[-1]                operand.pop()                while not s[i].isdigit():                    i += 1                i = self.findNextNumber(s, i, operand)                second = operand[-1]                operand.pop()                if s[current] == '*':                    operand.append(first * second)                else:                    if first < 0:                        first = (-1) * first                        x = (first / second) * (-1)                    else:                        x = first / second                    operand.append(x)                            result = operand[-1]        operand.pop()        while len(operand) != 0:            result += operand[-1]            operand.pop()        return result            def findNextNumber(self, s, pos, operand):        result = ord(s[pos]) - ord('0')        pos += 1        while pos < len(s) and (s[pos].isdigit() or s[pos] == ' '):            if s[pos] == ' ':                pos += 1                continue            else:                result = result * 10 + (ord(s[pos]) - ord('0'))                pos += 1        operand.append(result)        return pos