문제풀이 2021 01 12
2021.01.12 연습
- 사용언어 : java
class Solution {
public boolean isValid(String s) {
Stack<Character> stack = new Stack<Character>();
for (char i : s.toCharArray()) {
if ( i == '(' || i == '{' || i == '[') {
stack.push(i);
} else {
if (stack.isEmpty()) return false;
char check = stack.peek();
if ( i == ')' && check != '(' || i == '}' && check != '{' || i == ']' && check != '[' ) {
return false;
} else {
stack.pop();
}
}
}
return stack.isEmpty();
}
}
Runtime: 1 ms, faster than 98.74% of Java online submissions for Valid Parentheses. Memory Usage: 37.4 MB, less than 33.67% of Java online submissions for Valid Parentheses.
- 다른 코드 참고
class Solution {
// Hash table that takes care of the mappings.
private HashMap<Character, Character> mappings;
// Initialize hash map with mappings. This simply makes the code easier to read.
public Solution() {
this.mappings = new HashMap<Character, Character>();
this.mappings.put(')', '(');
this.mappings.put('}', '{');
this.mappings.put(']', '[');
}
public boolean isValid(String s) {
// Initialize a stack to be used in the algorithm.
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
// If the current character is a closing bracket.
if (this.mappings.containsKey(c)) {
// Get the top element of the stack. If the stack is empty, set a dummy value of '#'
char topElement = stack.empty() ? '#' : stack.pop();
// If the mapping for this bracket doesn't match the stack's top element, return false.
if (topElement != this.mappings.get(c)) {
return false;
}
} else {
// If it was an opening bracket, push to the stack.
stack.push(c);
}
}
// If the stack still contains elements, then it is an invalid expression.
return stack.isEmpty();
}
}