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.
先从最简单的开始练习:)
基本操作：遇到开始的括号，如 (, {, [进行压栈操作；遇到结束的括号，判断栈顶是否是匹配的括号), }, ]，如果不是或栈为空，则说明无效。最后判断栈是否为空即可。
利用了stack的特性，使用栈顶进行匹配动作。匹配就弹出，匹配就弹出……
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
switch (c) {
case '(':
stack.push('(');
break;
case '{':
stack.push('{');
break;
case '[':
stack.push('[');
break;
case ')':
if (stack.size() == 0 || stack.pop() != '(')
return false;
break;
case '}':
if (stack.size() == 0 || stack.pop() != '{')
return false;
break;
case ']':
if (stack.size() == 0 || stack.pop() != '[')
return false;
break;
}
}
return stack.isEmpty();
}
看了这里的答案，代码逻辑优化，算法是一样的。但的确优美了一些——
public boolean isValidSimpler(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();
}