Problem
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.
Note that an empty string is also considered valid.
Example 1:
Input: "()"
Output: true
Example 2:
Input: "()[]{}"
Output: true
Example 3:
Input: "(]"
Output: false
Example 4:
Input: "([)]"
Output: false
Example 5:
Input: "{[]}"
Output: true
Solution (Java)
class Solution {
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '(' || c == '[' || c == '{') {
stack.push(c);
} else if (c == ')' && !stack.isEmpty() && stack.peek() == '(') {
stack.pop();
} else if (c == '}' && !stack.isEmpty() && stack.peek() == '{') {
stack.pop();
} else if (c == ']' && !stack.isEmpty() && stack.peek() == '[') {
stack.pop();
} else {
return false;
}
}
return stack.isEmpty();
}
}
Solution (Javascript)
/**
* @param {string} s
* @return {boolean}
*/
var isValid = function(s) {
var stack = [];
var len = s.length;
var map = {
'(': ')',
'[': ']',
'{': '}'
};
for (var i = 0; i < len; i++) {
if (stack.length > 0 && map[stack[stack.length - 1]] === s[i]) {
stack.pop();
} else {
stack.push(s[i]);
}
}
return stack.length === 0;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).