LeetCode.224 Basic Calculator

字数统计: 416阅读时长: 2 min

 2019/06/04   Share

https://leetcode.com/problems/basic-calculator/

题意

实现一个带“（”，“）”，“+”，“-”运算符的简易计算器
解法

不难但是好复杂

class Solution {
  public int calculate(String s) {
      s = s.replaceAll("\\s*", "");
      int i = 0;
      Stack<String> stack = new Stack<>();
      while ( i < s.length()) {
          if (s.charAt(i) == '(') {
              //将(入栈
              stack.push("(");
              i++;
          } else if (s.charAt(i) == ')') {
              //将上一个结果出栈
              String current = stack.pop();
              //pop "("，弹出（
              stack.pop();
              //判断stack是否还有运算符和数字
              if (!stack.isEmpty()) {
                  String op = stack.pop();
                  String last = stack.pop();
                  if ("+".equals(op)) {
                      stack.push(String.valueOf(Integer.parseInt(last) + Integer.parseInt(current)));
                  } else {
                      stack.push(String.valueOf(Integer.parseInt(last) - Integer.parseInt(current)));
                  }
              } else {
                  //将当前结果入栈
                  stack.push(current);
              }
              i++;
          } else if (s.charAt(i) == '+') {
              //判断后一个是否是(，不是则计算与上一个的结果并入栈，否则将算符入栈
              if (s.charAt(i + 1) != '(') {
                  int end = getNextNumber(s, i + 1);
                  String current = s.substring(i+1, end);
                  String last = stack.pop();
                  stack.push(String.valueOf(Integer.parseInt(last) + Integer.parseInt(current)));
                  i = end;
              } else {
                  stack.push("+");
                  i+=1;
              }
          } else if (s.charAt(i) == '-') {
              if (s.charAt(i + 1) != '(') {
                  int end = getNextNumber(s, i + 1);
                  String current = s.substring(i+1, end);
                  String last = stack.pop();
                  stack.push(String.valueOf(Integer.parseInt(last) - Integer.parseInt(current)));
                  i = end;
              } else {
                  stack.push("-");
                  i+=1;
              }
          } else {
              int end = getNextNumber(s, i);
              //拿出当前数字
              String current = s.substring(i, end);
              //判断栈中是否还有其它算符
              if (!stack.isEmpty() && !"(".equals(stack.peek())) {
                  String op = stack.pop();
                  String last = stack.pop();
                  if ("+".equals(op)) {
                      stack.push(String.valueOf(Integer.parseInt(last) + Integer.parseInt(current)));
                  } else {
                      stack.push(String.valueOf(Integer.parseInt(last) - Integer.parseInt(current)));
                  }
              } else {
                  stack.push(current);
              }
              i = end;
          }
      }
      return Integer.parseInt(stack.pop());
  }

  public int getNextNumber(String s, int pos) {
      int end = pos;
      while (end < s.length() && s.charAt(end) >= '0' && s.charAt(end) <= '9') {
          end++;
      }
      return end;
  }
}

原文作者：Matteo

原文链接：http://tlzxsun.github.io/2019/06/04/LeetCode-224-Basic-Calculator/

发表日期：June 4th 2019, 4:46:38 pm

更新日期：June 4th 2019, 4:46:39 pm