[LeetCode] 20. Valid Parentheses 合法括号

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

An input string is valid if:

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

输入的字符串只含有'(', ')', '{', '}', '[' , ']'，验证字符串是否配对合理。

解法：栈Stack，经典的使用栈的题目。遍历字符串，如果为左括号，将其压入栈中，如果遇到为右括号，若此时栈为空，则直接返回false，如不为空，则取出栈顶元素，若为对应的左括号，是合法的继续循环，否则返回false。

Java: Time: O(n), Space: O(n)

public static boolean isValid(String s) {
    HashMap<Character, Character> map = new HashMap<Character, Character>();
    map.put('(', ')');
    map.put('[', ']');
    map.put('{', '}');
    Stack<Character> stack = new Stack<Character>();
  
    for (int i = 0; i < s.length(); i++) {
        char curr = s.charAt(i); 
        if (map.keySet().contains(curr)) {
            stack.push(curr);
        } else if (map.values().contains(curr)) {
            if (!stack.empty() && map.get(stack.peek()) == curr) {
                stack.pop();
            } else {
                return false;
            }
        }
    } 
    return stack.empty();
}

Java:

public boolean isValid(String s) {
    Stack<Character> stack = new Stack<Character>();
    for (char c : s.toCharArray()) {
        if (c == '(')
            stack.push(')');
        else if (c == '{')
            stack.push('}');
        else if (c == '[')
            stack.push(']');
        else if (stack.isEmpty() || stack.pop() != c)
            return false;
    }
    return stack.isEmpty();
}　　

Python:

class Solution:
    # @return a boolean
    def isValid(self, s):
        stack = []
        dict = {"]":"[", "}":"{", ")":"("}
        for char in s:
            if char in dict.values():
                stack.append(char)
            elif char in dict.keys():
                if stack == [] or dict[char] != stack.pop():
                    return False
            else:
                return False
        return stack == []　　

Python: Time: O(n), Space: O(n)

class Solution:
    def isValid(self, s):
        stack, lookup = [], {"(": ")", "{": "}", "[": "]"}
        for parenthese in s:
            if parenthese in lookup:  # Cannot use if lookup[parenthese]: KeyError
                stack.append(parenthese)
            elif len(stack) == 0 or lookup[stack.pop()] != parenthese: # Cannot use not stack
                return False
        return len(stack) == 0

Python: wo

class Solution(object):
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        stack = []
        m = {'(': ')', '[': ']', '{': '}'}
        n = [')', ']', '}']
        for i in s:
            if i in m:
                stack.append(i)
            elif i in n and stack:
                if m[stack.pop()] != i:
                    return False
            else:
                return False
        return len(stack) == 0

C++:

class Solution {
public:
    bool isValid(string s) {
        stack<char> parentheses;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == '(' || s[i] == '[' || s[i] == '{') parentheses.push(s[i]);
            else {
                if (parentheses.empty()) return false;
                if (s[i] == ')' && parentheses.top() != '(') return false;
                if (s[i] == ']' && parentheses.top() != '[') return false;
                if (s[i] == '}' && parentheses.top() != '{') return false;
                parentheses.pop();
            }
        }
        return parentheses.empty();
    }
}; 