【LeetCode】Basic Calculator 解题报告
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【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; }}
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