【leetcode系列】Valid Parentheses

標籤：leetcode   valid parentheses   

很經典的問題，使用棧來解決，我這裡自己實現了一個棧，當然也可以直接用java內建的Stack類。

自己實現的棧代碼：

import java.util.LinkedList;class StackOne {LinkedList<Object> data;int top;int maxSize;StackOne(int size) {// TODO Auto-generated constructor stubtop = -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("棧滿");}data.addLast(object);top++;return true;}Object pop() throws Exception {if (isEmpty()) {throw new Exception("棧空");}top--;return data.removeLast();}Object peek() {return data.getLast();}}

判斷輸出是否有效：

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 blocke.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 blocke.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 blocke.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 blocke.printStackTrace();}if (tmp == '{') {result = true;}}}}if (!stackOne.isEmpty()) {result = false;}return result;}public static void main(String[] args) {System.out.print(isValid("(}"));}}