【LeetCode】20. Valid Parentheses - Java实现

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1. 题目描述：

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. 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

2. 思路分析：

题目的意思是给定一个包含一系列括号的字符串，判断其中的括号是否两两匹配。

括号匹配问题，很自然地想到用栈来处理，即遍历字符串，遇到左括号就入栈，遇到右括号，则出栈并判断与当前的右括号是否匹配。

3. Java代码：

源代码：见我GiHub主页

代码：

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();
}