[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:

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

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();
    }
}; 

　　

类似题目：

[LeetCode] 22. Generate Parentheses

[LeetCode] 32. Longest Valid Parentheses

[LeetCode] 241. Different Ways to Add Parentheses

[LeetCode] 301. Remove Invalid Parentheses

 

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