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.
Example 1:
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| Input: s = "()"
Output: true
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Example 2:
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| Input: s = "()[]{}"
Output: true
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Example 3:
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| Input: s = "(]"
Output: false
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Example 4:
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| Input: s = "([)]"
Output: false
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Example 5:
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| Input: s = "{[]}"
Output: true
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Constraints:
1 <= s.length <= 104s consists of parentheses only '()[]{}'.
思路：
思路一：栈
在栈的那一节中，就提及到了栈的其中一个应用就是括号匹配，大致思路如下：
遇到 (、 {、 [ 这三种左括号，直接让它入栈，直到出现 ）、 }、 ] ，首先判断栈是否为空或者栈顶的元素是否对应，如果对应，那么让它出栈，继续上述操作即可
代码：
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| public boolean isValid(String s) {
if (s == null || s.length() == 0) return true;
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
char temp = s.charAt(i);
if (temp != ')' && temp != ']' && temp != '}') {
stack.push(s.charAt(i));
} else if (temp == ')' && !stack.isEmpty() && stack.peek() == '(' ||
temp == ']' && !stack.isEmpty() && stack.peek() == '[' ||
temp == '}' && !stack.isEmpty() && stack.peek() == '{') {
stack.pop();
} else {
return false;
}
}
return stack.isEmpty();
}
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时间复杂度：O(n)
空间复杂度：O(n)
还看到了一种反向写法
代码：
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| public boolean isValid(String s) {
if(s.isEmpty()) { return true; }
Stack<Character> stack = new Stack<>();
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();
}
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