978. Basic Calculator

https://www.lintcode.com/problem/basic-calculator/description

public class Solution {
/**
* @param s: the given expression
* @return: the result of expression
*/
public int calculate(String s) {
// Write your code here
Stack<Integer> stack = new Stack<Integer>();
int result =0;
int number =0;
int sign =1; for(int i =0;i<s.length();i++){
char c = s.charAt(i);
if(Character.isDigit(c)){
number = 10*number + (int)(c-'0');
}else if(c=='+'){
result +=sign*number;
number =0;
sign =1;
}else if(c == '-'){
result += sign * number;
number = 0;
sign = -1;
}else if(c == '('){
stack.push(result);
stack.push(sign);
sign = 1;
result = 0;
}else if(c == ')'){
result += sign* number;
number =0;
result*=stack.pop();
result+=stack.pop();
}
} if(number!=0){
result +=sign*number;
} return result;
}
}

980. Basic Calculator II

https://www.lintcode.com/problem/basic-calculator-ii/description

public class Solution {
/**
* @param s: the given expression
* @return: the result of expression
*/
public int calculate(String s) {
// Write your code here
if(s==null || s.length()==0){
return 0;
} int len = s.length();
Stack<Integer> stack = new Stack<Integer>();
int num =0;
char sign = '+';
for(int i =0;i<len;i++){ if(Character.isDigit(s.charAt(i))){
num = num*10 + s.charAt(i)-'0';
} if((!Character.isDigit(s.charAt(i)) && ' '!=s.charAt(i)) ||i==len-1){
if(sign =='-'){
stack.push(-num);
}
if(sign == '+'){
stack.push(num);
}
if(sign == '*'){
stack.push(stack.pop()*num);
}
if(sign == '/'){
stack.push(stack.pop()/num);
}
sign = s.charAt(i);
num =0;
}
} int re =0;
for(int i:stack){
re +=i;
}
return re; } }

849. Basic Calculator III

https://www.lintcode.com/problem/basic-calculator-iii/description

public class Solution {
/**
* @param s: the expression string
* @return: the answer
*/
public int calculate(String s) {
// Write your code here
if(s==null || s.length()==0){
return 0;
} Stack<Integer> nums = new Stack<Integer>();
Stack<Character> opr = new Stack<Character>(); int num =0;
for(int i=0;i<s.length();i++){
char c = s.charAt(i);
if(c==' '){
continue;
}
if(Character.isDigit(c)){
num = c-'0';
while(i<s.length()-1 && Character.isDigit(s.charAt(i+1))){
num = num*10+ (s.charAt(i+1)-'0');
i++;
}
nums.push(num);
num =0;
}else if(c=='('){
opr.push(c);
}else if(c==')'){
while(opr.peek()!='('){
nums.push(calculate(nums.pop(),nums.pop(),opr.pop()));
}
opr.pop();
}else if(c=='+'||c=='-'||c=='*'||c=='/'){
if(!opr.isEmpty()&&needCalFirst(opr.peek(),c)){
nums.push(calculate(nums.pop(),nums.pop(),opr.pop()));
}
opr.push(c);
}
} while(!opr.isEmpty()){
nums.push(calculate(nums.pop(),nums.pop(),opr.pop()));
} return nums.pop();
} public int calculate(int num1, int num2, char op){
switch(op){
case '+':return num2+num1;
case '-':return num2-num1;
case '*':return num2*num1;
case '/':return num2/num1;
default:return 0;
}
} public boolean needCalFirst(char firstOp, char secondOp){
if(firstOp=='('|| firstOp==')'){
return false;
} if((firstOp=='+'||firstOp=='-')&&(secondOp=='*'||secondOp=='/')){
return false;
} return true;
}
}

368. Expression Evaluation

https://www.lintcode.com/problem/expression-evaluation/description?_from=ladder&&fromId=4

同849 只需注意输入可能不是一个可计算表达式 eg:{'(',')'}

public class Solution {
/**
* @param expression: a list of strings
* @return: an integer
*/
public int evaluateExpression(String[] expression) {
// write your code here
if(expression==null || expression.length==0){
return 0;
} String s = "";
for(int i=0;i<expression.length;i++){
s+=expression[i];
} Stack<Integer> nums = new Stack<Integer>();
Stack<Character> opr = new Stack<Character>(); int num =0;
for(int i=0;i<s.length();i++){
char c = s.charAt(i);
if(c==' '){
continue;
}
if(Character.isDigit(c)){
num = c-'0';
while(i<s.length()-1 && Character.isDigit(s.charAt(i+1))){
num = num*10+ (s.charAt(i+1)-'0');
i++;
}
nums.push(num);
num =0;
}else if(c=='('){
opr.push(c);
}else if(c==')'){
while(opr.peek()!='('){
if(nums.size()>=2)
nums.push(calculate(nums.pop(),nums.pop(),opr.pop()));
}
opr.pop();
}else if(c=='+'||c=='-'||c=='*'||c=='/'){
if(!opr.isEmpty()&&needCalFirst(opr.peek(),c)){
if(nums.size()>=2)
nums.push(calculate(nums.pop(),nums.pop(),opr.pop()));
}
opr.push(c);
}
} while(!opr.isEmpty()){
if(nums.size()>=2)
nums.push(calculate(nums.pop(),nums.pop(),opr.pop()));
} if(nums.size()>0){
return nums.pop();
} return 0;
} public int calculate(int num1, int num2, char op){
switch(op){
case '+':return num2+num1;
case '-':return num2-num1;
case '*':return num2*num1;
case '/':return num2/num1;
default:return 0;
}
} public boolean needCalFirst(char firstOp, char secondOp){
if(firstOp=='('|| firstOp==')'){
return false;
} if((firstOp=='+'||firstOp=='-')&&(secondOp=='*'||secondOp=='/')){
return false;
} return true;
}
}

Basic Calculator - Stack(表达式计算器)的更多相关文章

  1. LeetCode OJ:Basic Calculator(基础计算器)

    Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...

  2. [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 ...

  3. [LeetCode] 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 ...

  4. [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 ...

  5. [LeetCode] Basic Calculator IV 基本计算器之四

    Given an expression such as expression = "e + 8 - a + 5" and an evaluation map such as {&q ...

  6. 227 Basic Calculator II 基本计算器II

    实现一个基本的计算器来计算一个简单的字符串表达式. 字符串表达式仅包含非负整数,+, - ,*,/四种运算符和空格 . 整数除法仅保留整数部分. 你可以假定所给定的表达式总是有效的. 一些例子: &q ...

  7. [LeetCode] 224. Basic Calculator 基本计算器

    Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...

  8. [LeetCode] Basic Calculator 基本计算器

    Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...

  9. [Swift]LeetCode224. 基本计算器 | Basic Calculator

    Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...

随机推荐

  1. 网络爬虫--requests库中两个重要的对象

    当我们使用resquests.get()时,返回的时response的对象,他包含服务器返回的所有信息,也包含请求的request的信息. 首先: response对象的属性有以下几个, r.stat ...

  2. ceph故障:too many PGs per OSD

    原文:http://www.linuxidc.com/Linux/2017-04/142518.htm 背景 集群状态报错,如下: # ceph -s cluster 1d64ac80-21be-43 ...

  3. Webservice初级问题: FAILED TO READ WSDL document

    这个问题是说明,这个版本的没法下载 犯错的图样 处理方法一: 将网页上xml文档下载,保存在本地,然后错误提示的这几行删除,保存文档,然后从本地调用 (1)右键另存为 保存为文件名a.xml (2)打 ...

  4. .NET基础 (12)时间的操作System.DateTime

    时间的操作System.DateTime1 DateTime如何存储时间2 如何在DateTime对象和字符串对象之间进行转换3 什么是UTC时间,如何转换到UTC时间 时间的操作System.Dat ...

  5. 自定义View--滚动View

    实现这么一个效果,一个布局中有一个View,那个View会随着我们手指的拖动而滑动,这种效果该如何实现?   我们第一反应应该是自定义一个DragView类继承View,然后重写onTouchEven ...

  6. handsontable-utilities

    搜索值 鼠标右键 讲了四个功能:1.row header是否可以右键(rowheader:true):2.删除右键列表的某些值(通过数组定义):3.自定义右键列表和功能(callback,item两个 ...

  7. URAL 1996 Cipher Message 3 (FFT + KMP)

    转载请注明出处,谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents    by---cxlove 题意 :给出两个串A , B,每个串是若干个byt ...

  8. 数据恢复软Extundelete

    1>概述  作为一名运维人员,保证数据的安全是根本职责,所以在维护系统的时候,要慎重和细心,但是有时也难免发生出现数据被误删除的情况,这个时候该如何              快速.有效地恢复数 ...

  9. CORS 跨域请求

    一.简介 CORS需要浏览器和服务器同时支持.目前,所有浏览器都支持该功能,IE浏览器不能低于IE10. 整个CORS通信过程,都是浏览器自动完成,不需要用户参与.对于开发者来说,CORS通信与同源的 ...

  10. Linux 连接数过多排查思路

    ## 在连接数报警的机器上,查看某个端口tcp连接来源,并排序 netstat -natl |grep ^tcp |grep ":2181" |awk '{print $5}'|a ...