LeetCode 20. Valid Parentheses

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博客讨论了判断包含特定括号字符的字符串是否有效的问题，需满足括号类型匹配且按正确顺序闭合，空字符串也视为有效。还介绍了 stack.peek() 与 stack.pop() 的区别，二者都返回栈顶值，但 peek() 不改变栈，pop() 会删除栈顶值。

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

stack.peek()与stack.pop()区别

相同点：大家都返回栈顶的值。

不同点：peek()不改变栈的值(不删除栈顶的值)，pop()会把栈顶的值删除。

//java
public static boolean isValid(String s) {
    Stack<Character> stack = new Stack<>();
    for (int i = 0; i < s.length(); i++) {
        char ch = s.charAt(i);
        // 如果是左括号，则入栈
        if (ch == '(' || ch == '[' || ch == '{') {
            stack.push(ch);
        } else { 
            // 如果是右括号，则比较其与栈顶元素是否配对
            if (stack.isEmpty()) {
                return false;
            }
            if (ch == ')' && stack.peek() != '(') {
                return false;
            }
            if (ch == ']' && stack.peek() != '[') {
                return false;
            }
            if (ch == '}' && stack.peek() != '{') {
                return false;
            }
            stack.pop();
        }
    }
    // 最后栈为空则表示完全匹配完毕
    return stack.isEmpty();
}