[leetcode] Valid Parentheses

最新推荐文章于 2019-07-31 23:36:17 发布

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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。通过遍历字符串中的每一种括号，确保它们能够按照正确的顺序闭合。具体实现包括两种情况：遇到开括号时将其压入栈中；遇到闭括号时检查栈顶元素是否为对应的开括号。

From : https://leetcode.com/problems/valid-parentheses

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) {
        stack<char> box;
        int len = s.size();
        char cur;
        for(int i=0; i<len; i++) {
            cur = s[i];
            if(cur=='(' || cur=='[' || cur=='{') box.push(cur);
            else if(cur==')' || cur==']' || cur=='}') {
                if(box.empty()) return false;
                char pre = box.top();
                switch(cur) {
                    case ')' : {
                        if(pre == '(') box.pop();
                        else return false;
                        break;
                    }
                    case ']' : {
                        if(pre == '[') box.pop();
                        else return false;
                        break;
                    }
                    case '}' : {
                        if(pre == '{') box.pop();
                        else return false;
                        break;
                    }
                }
            }
        }
        
        return box.empty();
    }
};

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