Leetcode 224: Basic Calculator

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Basic Calculator

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.

[思路]

遇到 ‘(‘ 就把之前的结果和符号push进stack. 遇到‘)‘就把 当前结果*stack中的符号 再加上stack中之前的结果.

[CODE]

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

Leetcode 224: Basic Calculator

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原文地址：http://blog.csdn.net/xudli/article/details/46554835