[LeetCode] 224. Basic Calculator 基本计算器
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.
Java:
public int calculate(String s) {
// delte white spaces
s = s.replaceAll(" ", "");
Stack<String> stack = new Stack<String>();
char[] arr = s.toCharArray();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
if (arr[i] == ' ')
continue;
if (arr[i] >= '0' && arr[i] <= '9') {
sb.append(arr[i]);
if (i == arr.length - 1) {
stack.push(sb.toString());
}
} else {
if (sb.length() > 0) {
stack.push(sb.toString());
sb = new StringBuilder();
}
if (arr[i] != ')') {
stack.push(new String(new char[] { arr[i] }));
} else {
// when meet ')', pop and calculate
ArrayList<String> t = new ArrayList<String>();
while (!stack.isEmpty()) {
String top = stack.pop();
if (top.equals("(")) {
break;
} else {
t.add(0, top);
}
}
int temp = 0;
if (t.size() == 1) {
temp = Integer.valueOf(t.get(0));
} else {
for (int j = t.size() - 1; j > 0; j = j - 2) {
if (t.get(j - 1).equals("-")) {
temp += 0 - Integer.valueOf(t.get(j));
} else {
temp += Integer.valueOf(t.get(j));
}
}
temp += Integer.valueOf(t.get(0));
}
stack.push(String.valueOf(temp));
}
}
}
ArrayList<String> t = new ArrayList<String>();
while (!stack.isEmpty()) {
String elem = stack.pop();
t.add(0, elem);
}
int temp = 0;
for (int i = t.size() - 1; i > 0; i = i - 2) {
if (t.get(i - 1).equals("-")) {
temp += 0 - Integer.valueOf(t.get(i));
} else {
temp += Integer.valueOf(t.get(i));
}
}
temp += Integer.valueOf(t.get(0));
return temp;
}
Python:
class Solution:
# @param {string} s
# @return {integer}
def calculate(self, s):
operands, operators = [], []
operand = ""
for i in reversed(xrange(len(s))):
if s[i].isdigit():
operand += s[i]
if i == 0 or not s[i-1].isdigit():
operands.append(int(operand[::-1]))
operand = ""
elif s[i] == ')' or s[i] == '+' or s[i] == '-':
operators.append(s[i])
elif s[i] == '(':
while operators[-1] != ')':
self.compute(operands, operators)
operators.pop() while operators:
self.compute(operands, operators) return operands[-1] def compute(self, operands, operators):
left, right = operands.pop(), operands.pop()
op = operators.pop()
if op == '+':
operands.append(left + right)
elif op == '-':
operands.append(left - right)
C++:
class Solution {
public:
int calculate(string s) {
int res = 0, sign = 1, n = s.size();
stack<int> st;
for (int i = 0; i < n; ++i) {
char c = s[i];
if (c >= '0') {
int num = 0;
while (i < n && s[i] >= '0') {
num = 10 * num + s[i++] - '0';
}
res += sign * num;
--i;
} else if (c == '+') {
sign = 1;
} else if (c == '-') {
sign = -1;
} else if (c == '(') {
st.push(res);
st.push(sign);
res = 0;
sign = 1;
} else if (c == ')') {
res *= st.top(); st.pop();
res += st.top(); st.pop();
}
}
return res;
}
};
C++:
class Solution2 {
public:
int calculate(string s) {
stack<int> operands;
stack<char> operators;
string operand;
for (int i = s.length() - 1; i >= 0; --i) {
if (isdigit(s[i])) {
operand.push_back(s[i]);
if (i == 0 || !isdigit(s[i - 1])) {
reverse(operand.begin(), operand.end());
operands.emplace(stoi(operand));
operand.clear();
}
} else if (s[i] == ')' || s[i] == '+' || s[i] == '-') {
operators.emplace(s[i]);
} else if (s[i] == '(') {
while (operators.top() != ')') {
compute(operands, operators);
}
operators.pop();
}
}
while (!operators.empty()) {
compute(operands, operators);
}
return operands.top();
}
void compute(stack<int>& operands, stack<char>& operators) {
const int left = operands.top();
operands.pop();
const int right = operands.top();
operands.pop();
const char op = operators.top();
operators.pop();
if (op == '+') {
operands.emplace(left + right);
} else if (op == '-') {
operands.emplace(left - right);
}
}
};
类似题目:
[LeetCode] 227. Basic Calculator II 基本计算器 II
All LeetCode Questions List 题目汇总
[LeetCode] 224. Basic Calculator 基本计算器的更多相关文章
- leetcode 224. Basic Calculator 、227. Basic Calculator II
这种题都要设置一个符号位的变量 224. Basic Calculator 设置数值和符号两个变量,遇到左括号将数值和符号加进栈中 class Solution { public: int calcu ...
- [leetcode]224. Basic Calculator
Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...
- Java for LeetCode 224 Basic Calculator
Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...
- (medium)LeetCode 224.Basic Calculator
Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...
- 224 Basic Calculator 基本计算器
实现一个基本的计算器来计算一个简单的字符串表达式. 字符串表达式可以包含左括号 ( ,右括号),加号+ ,减号 -,非负整数和空格 . 假定所给的表达式语句总是正确有效的. 例如: "1 + ...
- [LeetCode] 227. Basic Calculator II 基本计算器 II
Implement a basic calculator to evaluate a simple expression string. The expression string contains ...
- [LeetCode] Basic Calculator 基本计算器
Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...
- [LeetCode] 772. Basic Calculator III 基本计算器之三
Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...
- 【LeetCode】224. Basic Calculator 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 栈 参考资料 日期 题目地址:https://lee ...
随机推荐
- python正则表达式练习题
# coding=utf-8 import re # 1. 写一个正则表达式,使其能同时识别下面所有的字符串:'bat','bit', 'but', 'hat', 'hit', 'hut' s =&q ...
- unicode转换为中文
unicode转换为中文 \u5f53\u5730\u65f6\u95f42019\u5e747\u670813\u65e5\uff0c\u82f1\u56fd\u8d1d\u5fb7\u798f\u ...
- 解决Invalid character found in the request target. The valid characters are defined in RFC 7230 and RF
通过这里的回答,我们可以知道: Tomcat在 7.0.73, 8.0.39, 8.5.7 版本后,添加了对于http头的验证. 具体来说,就是添加了些规则去限制HTTP头的规范性 参考这里 具体来说 ...
- Visual Studio Code 写Python代码
之前用nodepad++,sublime text3,ultraedit,最近上手微软的vsc感觉上手还行,如果没有pycharm照样可以使用它 https://code.visualstudio.c ...
- 各大公司Java面试题收录含答案(整理版)持续中....
本文分为17个模块,分别是:Java基础.容器.多线程.反射.对象拷贝.Java web.异常.网络.设计模式.算法.Spring/Spring MVC.Spring Boot/Spring Clou ...
- zabbix4.2.5常见问题指南
一.zabbix配置postgres监控 rpm -ivh https://download.postgresql.org/pub/repos/yum/9.5/redhat/rhel-7-x86_64 ...
- c#的参数调用
c#的参数传递有三种方式:值传递,和c一样,引用传递,类似与c++,但形式不一样输出参数,这种方式可以返回多个值,这种有点像c中的指针传递,但其实不太一样.值传递不细说,c中已经很详细了引用传递实例如 ...
- gin+redis
var RedisDefaultPool *redis.Pool func newPool(addr string) *redis.Pool { return &redis.Pool{ Max ...
- 洛谷P1353 USACO 跑步 Running
题目 一道入门的dp,首先要先看懂题目要求. 容易得出状态\(dp[i][j]\)定义为i时间疲劳度为j所得到的最大距离 有两个坑点,首先疲劳到0仍然可以继续疲劳. 有第一个方程: \(dp[i][0 ...
- mysql upper() 函数
mysql> select upper(" cdcdcd"); +------------------+ | upper(" cdcdcd") | +-- ...