LeetCode: Valid Parentheses

最新推荐文章于 2022-02-25 19:38:17 发布

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最新推荐文章于 2022-02-25 19:38:17 发布

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分类专栏： LeetCode 文章标签： string input c

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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。通过三种不同类型的括号——圆括号、方括号及花括号，演示了如何判断输入字符串中的括号是否有效闭合。代码示例包括了初始化栈、遍历字符串、检查括号匹配情况等关键步骤。

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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!

#include<iostream>
#include<vector>
#include<string>
using namespace::std;



class Solution {
public:
    bool isValid(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function

		vector<char> tmp;
		int a = 0, b = 0, c = 0;
		char last;
		for(int i = 0; i < s.size(); i++)
		{
			
			switch(s[i])
			{
				case '(':
					tmp.push_back(s[i]);
					a++;
					break;
				case '{':
					tmp.push_back(s[i]);
					b++;
					break;
				case '[':
					tmp.push_back(s[i]);
					c++;
					break;
				case ')':
					if( last == '(' && a != 0)
					{
						a--;
						tmp.pop_back();
					}
					else
						return false;
					break;
				case '}':
					if( last == '{' && b != 0)
					{
						b--;
						tmp.pop_back();
					}
					else
						return false;
					break;
				case ']':
					if( last == '[' && c != 0)
					{
						c--;
						tmp.pop_back();
					}
					else
						return false;
					break;
			}
			if(tmp.size() != 0)
			last = tmp.back();

		}

		if(a != 0 || b != 0 || c != 0)
			return false;
		return true;


        
    }
};

int main()
{
	Solution ss;
	bool result = ss.isValid("((");
	cout<<"result:"<<result<<endl;
}

Round 2:

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

Round 3:

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