Leecode-Valid Parentheses

最新推荐文章于 2024-08-05 09:52:52 发布

原创最新推荐文章于 2024-08-05 09:52:52 发布 · 198 阅读

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CC 4.0 BY-SA版权

本文介绍了一种使用栈数据结构来验证括号是否正确配对的方法。通过遍历字符串中的每一种括号类型，并利用栈来记录左括号，当遇到右括号时检查栈顶元素是否为对应的左括号，以此确保括号闭合顺序正确。

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

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

class Solution {
public:
    bool isValid(string s) {
	int length = s.size();
	bool validstring = false;
	std::stack<int> latests;

	for (int i = 0; i < length; i++)
	{
		switch (s[i]){
		case '(':
			latests.push(1);
			break;
		case '[':
			latests.push(2);
			break;
		case '{':
			latests.push(3);
			break;
		case ')':
			if (latests.empty())
				return validstring = false;
			if (latests.top() == 1)
			{
				latests.pop();
				break;
			}
			else
				validstring = false;
			return validstring;
		case ']':
			if (latests.empty())
				return validstring = false;
			if (latests.top() == 2)
			{
				latests.pop();
				break;
			}
			else
				validstring = false;
			return validstring;
		case '}':
			if (latests.empty())
				return validstring = false;
			if (latests.top() == 3)
			{
				latests.pop();
				break;
			}
			else
				validstring = false;
			return validstring;
		}
	}
	if (latests.empty())//空栈证明所有前括号都有结尾，栈顺序保证符号顺序
	{
		validstring = true;
	}
	return validstring;
    }
};

bool isValid(string s) {
	int length = s.size();
	bool validstring = false;
	std::stack<int> latests;


	for (int i = 0; i < length; i++)
	{
		switch (s[i]){
		case '(':
			latests.push(1);
			break;
		case '[':
			latests.push(2);
			break;
		case '{':
			latests.push(3);
			break;
		case ')':
			if (latests.empty()||latests.top() != 1)
				return validstring = false;
			latests.pop();
			break;
		case ']':
			if (latests.empty() || latests.top() != 2)
				return validstring = false;
			latests.pop();
			break;
		case '}':
			if (latests.empty() || latests.top() != 3)
				return validstring = false;
			latests.pop();
			break;
		}
	}
	if (latests.empty())
	{
		validstring = true;
	}
	return validstring;
}