【LeetCode】Basic Calculator 解题报告

标签：basic calculator   stack   leetcode   

【题目】

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】

直接拿测试用例走一下下面的程序，就会明吧其巧妙之处。

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;
    }
}

【LeetCode】Basic Calculator 解题报告

标签：basic calculator   stack   leetcode   

原文地址：http://blog.csdn.net/ljiabin/article/details/46489997