Valid parentheses (Stack)

Given A string containing just the characters,,, ‘(‘ ‘)‘ , and ‘{‘ ‘}‘ ‘[‘ ‘]‘ , determine if the input string I S valid.

The brackets must close in the correct order, and is all valid but and is not "()" "()[]{}" "(]" "([)]" .

public class Solution {

if (s = = null) return false; stack<character> stack = new stack<character> (); char[] ch = s.tochararray (); for (int i = 0; i < ch.length; i++) {if (ch[i] = = ' (') | | (Ch[i] = = ' [') | | (Ch[i] = = ' {')) Stack.push (Ch[i]), else {if (Stack.isempty ()) return false;if (Ch[i]-Stack.pop () > 2) return false;} return Stack.isempty ();

　　

    public Boolean isValid (String s) {        if (s = = null)                       //Improved 1: Can be used without string.charat (i)
                                   return false;          Improvement 2: Judge length first
                                              Improvement Three: Do not need so many if else, the symbol can be judged by ASCII code stack<character> stack = new stack<character> (); char[] ch = S.tochararray (); for (int i = 0; i < ch.length; i++) {if (ch[i] = = ' (') | | (Ch[i] = = ' [') | | (Ch[i] = = ' {')) Stack.push (Ch[i]), else {if (ch[i] = = ') ') {if (Stack.isempty ()) return False;else if (Stack.pop ()! = ' (') RET Urn false;} if (ch[i] = = '] ') {if (Stack.isempty ()) Return false;if (Stack.pop ()! = ' [') return false;} if (ch[i] = = '} ') {if (Stack.isempty ()) Return false;if (Stack.pop ()! = ' {') return false;}}} if (!stack.isempty ()) return False;elsereturn true;}    }

　　

Valid parentheses (Stack)