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:
public boolean isValid(String s) {
if(s==null|s.length()==0) return true;
char[] sc=s.toCharArray();
Stack<Character> path=new Stack<>();
for(char c:sc){
if(c=='('||c=='['||c=='{')
path.push(c);
else if(path.size()>0){
if(c==')'&&path.peek()=='('){
path.pop();
}else if(c==']'&&path.peek()=='['){
path.pop();
}else if(c=='}'&&path.peek()=='{'){
path.pop();
}else{
return false;
}
} else {
return false;
}
}
if(path.size()==0)
return true;
else
return false;
}
时间复杂度：O(n)
空间复杂度：O(n)
比较基础，只是题目中的要求基本上Example都给的很明确了，栈不为空，说明不匹配，栈为空时，第一个字符不是左括号的说明不匹配。其实还有一种简单的方法，别人写的，很灵性。
public boolean isValid(String s) {
Stack<Character> stack = new Stack<Character>();
for (char c : s.toCharArray()) {
if (c == '(')
stack.push(')');
else if (c == '{')
stack.push('}');
else if (c == '[')
stack.push(']');
else if (stack.isEmpty() || stack.pop() != c)
return false;
}
return stack.isEmpty();
}
思路就是只入后括号，再弹出。
测试用例
Case 1:
[
false
Case 2:
]
false
