题目介绍
Given a string s 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.
示例
Example 1:
Input: s = “()”
Output: true
Example 2:
Input: s = “()[]{}”
Output: true
Example 3:
Input: s = “(]”
Output: false
Constraints:
1 <= s.length <= 10^4s consists of parentheses only '()[]{}'.
解析
easy题，并且看起来也是比较简单的，三种括号按对匹配，直接用栈来做，栈里面存的是括号的类型，如果是左括号，就放入栈中，如果是右括号，就把栈顶的元素弹出，如果弹出的元素不是左括号，就返回false，如果弹出的元素是左括号，就继续往下走，如果遍历完了，如果栈里面还有元素，就返回false，如果遍历完了，如果栈里面没有元素，就返回true
代码
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| class Solution {
public boolean isValid(String s) {
if (s.length() % 2 != 0) {
return false;
}
Stack<String> stk = new Stack<>();
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '{' || s.charAt(i) == '(' || s.charAt(i) == '[') {
stk.push(String.valueOf(s.charAt(i)));
continue;
}
if (s.charAt(i) == '}') {
if (stk.isEmpty()) {
return false;
}
String cur = stk.peek();
if (cur.charAt(0) != '{') {
return false;
} else {
stk.pop();
}
continue;
}
if (s.charAt(i) == ']') {
if (stk.isEmpty()) {
return false;
}
String cur = stk.peek();
if (cur.charAt(0) != '[') {
return false;
} else {
stk.pop();
}
continue;
}
if (s.charAt(i) == ')') {
if (stk.isEmpty()) {
return false;
}
String cur = stk.peek();
if (cur.charAt(0) != '(') {
return false;
} else {
stk.pop();
}
continue;
}
}
return stk.size() == 0;
}
}
|