Created
January 27, 2013 22:34
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Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
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public class Solution { | |
public boolean isValid(String s) { | |
int length=s.length(); | |
char [] array=s.toCharArray(); | |
if(length==0) return true; | |
Stack<Character> stack=new Stack<Character>(); | |
for(int i=0; i<length; i++) | |
{ | |
if(array[i]=='(' || array[i]=='{' || array[i]=='[' ) | |
{ | |
stack.push(array[i]); | |
} | |
if(array[i]==')' ||array[i]=='}' ||array[i]==']') | |
{ | |
if(stack.isEmpty()) return false; | |
char temp=stack.pop(); | |
if((temp=='(' && array[i]==')' ) || (temp=='{' && array[i]=='}' ) ||(temp=='[' && array[i]==']' ) ) | |
{ | |
continue; | |
}else | |
{ | |
return false; | |
} | |
} | |
} | |
if(!stack.isEmpty()) return false; | |
return true; | |
} | |
} |
how about this?
const isValid = function (s: string) {
let isTrue = true;
if (s.length % 2 === 0) {
let currentPointer = 0;
for (let i = 0; i < s.length; i++) {
let selectedPair = s.substring(currentPointer, currentPointer + 2);
if (
(selectedPair.length > 1 && selectedPair === '()') ||
selectedPair === '{}' || selectedPair === '[]'
) {
if (currentPointer + 4 <= s.length) {
currentPointer = currentPointer + 2;
}
} else {
isTrue = false;
}
}
} else {
isTrue = false;
}
console.log(isTrue);
};
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easy without extra space stack or queue but with 2 pointers ( pre,next)