LeetCode 有效的括号

有效的括号

package leetcode;

import java.util.HashMap;
import java.util.Map;
import java.util.Stack;

/**
 * 给定一个只包括 '('，')'，'{'，'}'，'['，']' 的字符串，判断字符串是否有效。
 * 有效字符串需满足：
 * 左括号必须用相同类型的右括号闭合。
 * 左括号必须以正确的顺序闭合。
 * 注意空字符串可被认为是有效字符串。
 * 示例 1:
 * 输入: "()"
 * 输出: true
 * 示例 2:
 * 输入: "()[]{}"
 * 输出: true
 * 示例 3:
 * 输入: "(]"
 * 输出: false
 * 示例 4:
 * 输入: "([)]"
 * 输出: false
 * 示例 5:
 * 输入: "{[]}"
 * 输出: true
 */
public class MyTest {


    private HashMap<Character, Character> mappings;

    // Initialize hash map with mappings. This simply makes the code easier to read.
    public MyTest() {
        this.mappings = new HashMap<Character, Character>();
        this.mappings.put(')', '(');
        this.mappings.put('}', '{');
        this.mappings.put(']', '[');
    }

    public static void main(String[] args) {
        String s = "(]";
        //System.out.println(new MyTest().isValid(s));
        System.out.println(new MyTest().isValidNB(s));
    }


    /**
     * 笨比算法
     * @param s
     * @return
     */
    public boolean isValid(String s) {
        String str1 = "()";
        String str2 = "[]";
        String str3 = "{}";
        while (s.contains(str1) || s.contains(str2) || s.contains(str3)) {
            s = s.replace(str1, "").replace(str2, "").replace(str3, "");
        }
        return s.length()==0;
    }

    /**
     * 大神算法
     * 作者：LeetCode
     * 链接：https://leetcode-cn.com/problems/two-sum/solution/you-xiao-de-gua-hao-by-leetcode/
     * 来源：力扣（LeetCode）
     * 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
     *
     * @param s
     * @return
     */
    public boolean isValidNB(String s) {
        // Initialize a stack to be used in the algorithm.
        Stack<Character> stack = new Stack<Character>();

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);

            // If the current character is a closing bracket.
            if (this.mappings.containsKey(c)) {

                // Get the top element of the stack. If the stack is empty, set a dummy value of '#'
                char topElement = stack.empty() ? '#' : stack.pop();

                // If the mapping for this bracket doesn't match the stack's top element, return false.
                if (topElement != this.mappings.get(c)) {
                    return false;
                }
            } else {
                // If it was an opening bracket, push to the stack.
                stack.push(c);
            }
        }

        // If the stack still contains elements, then it is an invalid expression.
        return stack.isEmpty();
    }
}