问题
https://leetcode.com/problems/valid-parentheses/description/
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:
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Input: "()"
Output: true
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Example 2:
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Input: "()[]{}"
Output: true
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思路
使用Stack后进先出的特点，对于入栈的符号查看栈顶是否配对，配对则移除栈顶元素，否则入栈，最终查看栈是否为空，有两种实现方式。
代码
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public boolean (String s) {
if (s == null || s.length() == 0) {
return true;
}
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i ++) {
Character topChar = stack.isEmpty() ? null : stack.peek();
char curChar = s.charAt(i);
if (topChar != null) {
if (topChar.charValue() == '(') {
if (curChar == ')') {
stack.pop();
continue;
}
} else if (topChar.charValue() == '{') {
if (curChar == '}') {
stack.pop();
continue;
}
} else if (topChar.charValue() == '[') {
if (curChar == ']') {
stack.pop();
continue;
}
}
}
stack.push(Character.valueOf(curChar));
}
return stack.isEmpty();
}
public boolean isValid2(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();
}
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