【LeetCode】Basic Calculator 解題報告，leetcodecalculator

【LeetCode】Basic Calculator 解題報告，leetcodecalculator

【題目】

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

【解析】

直接拿測試案例走一下下面的程式，就會明吧其巧妙之處。

public class Solution {    public int calculate(String s) {        Stack<Integer> stack = new Stack<>();        stack.push(1);        stack.push(1);        int res = 0;        for (int i = 0; i < s.length(); i++) {            char c = s.charAt(i);            if (Character.isDigit(c)) {                int num = c - '0';                int j = i + 1;                while (j < s.length() && Character.isDigit(s.charAt(j))) {                    num = 10 * num + (s.charAt(j) - '0');                    j++;                }                res += stack.pop() * num;                i = j - 1;            } else if (c == '+' || c == '(') {                stack.push(stack.peek());            } else if (c == '-') {                stack.push(-1 * stack.peek());            } else if (c == ')') {                stack.pop();            }        }        return res;    }}