20. Title Leetcode effective brackets (simple)

Subject description:

Includes only a given '(', ')' string, '{', '}', '[', ']', and determine whether the string is valid.

Valid string must meet:

Left bracket must be closed by a closing parenthesis of the same type.
Left parenthesis must be closed in the correct order.
Note the empty string can be considered a valid string.

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

Topic Analysis:

LinkedList use as an auxiliary stack:

• Traversing the input string
• If the current character to the left half of the brackets, it is pushed onto the stack
• If you come across the right half of the brackets, Category talk:
• 1) If the stack is not empty and the left half of the corresponding bracket, the top element is removed, continue to cycle
• 2) If the stack is empty at this time, the process directly returns false
• 3) If the left half of the corresponding bracket, false otherwise

Code:

package com.company;

import java.util.LinkedList;

public class Main {

    public static void main(String[] args) {
        String str = "]";
        System.out.println(isValid(str));
    }

    private static boolean isValid(String s) {
        if (s == null || s.isEmpty()) {
            return true;
        }
        LinkedList<Character> linkedList = new LinkedList<>();
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (ch == '(' || ch == '[' || ch == '{') {
                linkedList.add(ch);
                // no element in the stack, and the current symbol is a right parenthesis, directly returns false, the current symbol does not need to push
            The else {}
                if (linkedList.isEmpty()) {
                    return false;
                }
                if (ch == ')') {
                    if (linkedList.peekLast() != '(') {
                        return false;
                    }
                }
                if (ch == ']') {
                    if (linkedList.peekLast() != '[') {
                        return false;
                    }
                }
                if (ch == '}') {
                    if (linkedList.peekLast() != '{'){
                        return false;
                    }
                }
                if (!linkedList.isEmpty()) {
                    linkedList.removeLast();
                }
            }
        }
        return linkedList.isEmpty();
    }
}

Time complexity: O (N), traversed again, n is the size of length of str

Space complexity: O (N), LinkedList size n, i.e. the auxiliary stack size