Update validParentheses.java

fix code logic error leading to incorrect solution: top of stack should be an opening parenthesis if current character is a closing parenthesis for string to be valid
zhijunxie · Apr 10, 2017 · 7b83827 · 7b83827
1 parent 9ab343f
commit 7b83827
Showing 1 changed file with 4 additions and 4 deletions.
diff --git a/Company/Airbnb/validParentheses.java b/Company/Airbnb/validParentheses.java
@@ -22,19 +22,19 @@ public boolean isValid(String s) {
 
             }
 
-            else if(s.charAt(i) == ')' && !stack.isEmpty() && stack.peek() == ')') {
+            else if(s.charAt(i) == ')' && !stack.isEmpty() && stack.peek() == '(') {
 
                 stack.pop();
 
             }
 
-            else if(s.charAt(i) == ']' && !stack.isEmpty() && stack.peek() == ']') {
+            else if(s.charAt(i) == ']' && !stack.isEmpty() && stack.peek() == '[') {
 
                 stack.pop();
 
             }
 
-            else if(s.charAt(i) == '}' && !stack.isEmpty() && stack.peek() == '}') {
+            else if(s.charAt(i) == '}' && !stack.isEmpty() && stack.peek() == '{') {
 
                 stack.pop();
 
@@ -52,4 +52,4 @@ else if(s.charAt(i) == '}' && !stack.isEmpty() && stack.peek() == '}') {
 
     }
 
-}
+}