python brush LeetCode:. 20 effective brackets

Difficulty level: Easy

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

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/valid-parentheses
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Problem-solving ideas:

1, using the stack thought, traversing the left parenthesis added to the list

2. If you encounter a right parenthesis, and a pre-match and match brackets

3. If the match, that delete the brackets match, continue to traverse

Problem-solving Code:

class Solution:
     DEF isValid (Self, S: STR) -> BOOL: 
        S = List (S)   # into a list, because the string is immutable 

        left_dics = []   # for storing the left bracket 
        Rights = [ ' ) ' , ' ] ' , ' } ' ] 
        paires = {
             ' ( ' : ' ) ' ,
             ' [ ' : ' ] ' ,
             ' {' : ' } ' 
        } 

        For char in S:
             IF char in paires: 
                left_dics.append (char)   # If left parenthesis, joined list 
            elif char in Rights and left_dics:
                 IF char == paires.get (left_dics [- 1]):   # if the brackets on the right and in front of a relatively parenthesis match 
                    left_dics.pop ()   # if you delete the original list of matching last brace 
                the else :
                     BREAK 
            the else : 
                left_dics = 1   #If not match, out of the loop, this case is primarily exclude the first element of the anti-brackets situation 
                BREAK 

        IF  not left_dics:
             return True
         the else :
             return False