【leetcode系列】Valid Parentheses

很经典的问题，使用栈来解决，我这里自己实现了一个栈，当然也可以直接用java自带的Stack类。

自己实现的栈代码：

import java.util.LinkedList;

class StackOne {
LinkedList<Object> data;
int top;
int maxSize;

StackOne(int size) {
// TODO Auto-generated constructor stub
top = -1;
maxSize = 100;
data = new LinkedList<Object>();
}

int getElementCount() {
return data.size();
}

boolean isEmpty() {
return top == -1;
}

boolean isFull() {
return top + 1 == maxSize;
}

boolean push(Object object) throws Exception {
if (isFull()) {
throw new Exception("栈满");
}
data.addLast(object);
top++;
return true;
}

Object pop() throws Exception {
if (isEmpty()) {
throw new Exception("栈空");
}
top--;
return data.removeLast();
}

Object peek() {
return data.getLast();
}
}


判断输出是否有效：
public class Solution {

public static boolean isValid(String in) {
StackOne stackOne = new StackOne(100);
boolean result = false;
char[] inArray = in.toCharArray();
for (char i : inArray) {
if (i == '(' || i == '[' || i == '{') {
try {
stackOne.push(i);
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
continue;
}
if (i == ')') {
if (stackOne.isEmpty()) {
result = false;
} else {
char tmp = '\u0000';
try {
tmp = (Character) stackOne.pop();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
if (tmp == '(') {
result = true;
}
}
}
if (i == ']') {
if (stackOne.isEmpty()) {
result = false;
} else {
char tmp = '\u0000';
try {
tmp = (Character) stackOne.pop();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
if (tmp == '[') {
result = true;
}
}
}
if (i == '}') {
if (stackOne.isEmpty()) {
result = false;
} else {
char tmp = '\u0000';
try {
tmp = (Character) stackOne.pop();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
if (tmp == '{') {
result = true;
}
}

}
}
if (!stackOne.isEmpty()) {
result = false;
}
return result;
}

public static void main(String[] args) {
System.out.print(isValid("(}"));
}
}