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    • LeetCode Solutions
      • [LeetCode] 1. Two Sum [Easy]
      • [LeetCode] 2. Add Two Numbers [Medium]
      • [LeetCode] 3. Longest Substring Without Repeating Characters [Medium]
      • [LeetCode] 5. Longest Palindromic Substring [Medium]
      • [LeetCode] 7. Reverse Integer [Easy] [LeetCode]
      • [LeetCode] 8. String to Integer (atoi)
      • [LeetCode] 11. Container With Most Water
      • [LeetCode] 13. Roman to Integer
      • [LeetCode]14. Longest Common Prefix
      • [LeetCode] 15. 3Sum
      • [LeetCode] 17. Letter Combinations of a Phone Number
      • [LeetCode] 19. Remove Nth Node From End of List
      • [LeetCode] 20. Valid Parentheses
      • [LeetCode] 21. Merge Two Sorted Lists
      • [LeetCode] 22. Generate Parentheses
      • [LeetCode] 26. Remove Duplicates from Sorted Array
      • [LeetCode] 28. Implement strStr()
      • [LeetCode] 33. Search in Rotated Sorted Array
      • [LeetCode] 34. Find First and Last Position of Element in Sorted Array
      • [LeetCode]36. Valid Sudoku
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      • [LeetCode] 46. Permutations
      • [LeetCode] 48. Rotate Image
      • [LeetCode] 49. Group Anagrams
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  • About Author
    • Frank Chen
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  1. Leet Code
  2. LeetCode Solutions

[LeetCode] 20. Valid Parentheses

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

Analyse:

​

Solution:

class Solution {
    func isValid(_ s: String) -> Bool {
          let leftChars:[Character] = ["(", "{", "["];
        let rightChars:[Character] = [")", "}", "]"];
        
        var stack:[Character] = [];
        let chars = Array(s);
        
        
        if(chars.count % 2 != 0) {
            return false;
        }
        
        for i in 0..<chars.count {
            let char = chars[i];
            
            if(leftChars.index(of:char) != nil) {
                stack.append(char);
            }
            else if(rightChars.index(of:char) != nil) {
                
                if(stack.count == 0) {
                    return false;
                }
                   
                if(leftChars.index(of:stack.last!) == rightChars.index(of:char)) {
                    stack.removeLast();
                }
                else {
                    return false;
                }
            }
        }
        
        if(stack.count == 0) {
            return true;
        }
        
        return false;
    }
}
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Last updated 6 years ago