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validParenthesis.java
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Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if:
Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
//O(n) to go through length of string
//O(n) because stack could have length n in worst case
class Solution {
public boolean isValid(String s) {
Stack<Character> stack = new Stack<Character>();
// Iterate through string until empty
for(int i = 0; i<s.length(); i++) {
// Push any open parentheses onto stack
if(s.charAt(i) == '(' || s.charAt(i) == '[' || s.charAt(i) == '{')
stack.push(s.charAt(i));
// Check stack for corresponding closing parentheses, false if not valid
else if(s.charAt(i) == ')' && !stack.empty() && stack.peek() == '(')
stack.pop();
else if(s.charAt(i) == ']' && !stack.empty() && stack.peek() == '[')
stack.pop();
else if(s.charAt(i) == '}' && !stack.empty() && stack.peek() == '{')
stack.pop();
else
return false;
}
// return true if no open parentheses left in stack
return stack.empty();
}
}
//MOST BASIC WAY TO IMPLEMENT
class Solution {
public boolean isValid(String s) {
if(s == null) return true;
Map<Character, Character> map = new HashMap<>();
map.put(')', '(');
map.put('}','{');
map.put(']', '[');
Stack<Character> stack = new Stack<>();
for(char c: s.toCharArray()){
if(c == '(' || c == '[' || c=='{'){
stack.push(c);
} else {
if(!stack.isEmpty() && stack.peek() != map.get(c)){
return false;
} else{
if(!stack.isEmpty())
{
stack.pop();
} else{
return false;
}
}
}
}
if(stack.size()!=0) return false;
return true;
}
}