题目:
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.
代码1:
class Solution { public: bool isValid(string s) { stack<char> bracket; for(int i=0;i<s.size();i++){ if(s[i]=='('||s[i]=='['||s[i]=='{'){ bracket.push(s[i]); }else if(s[i]==')'&&!bracket.empty()&&bracket.top()=='('){ bracket.pop(); }else if(s[i]==']'&&!bracket.empty()&&bracket.top()=='['){ bracket.pop(); }else if(s[i]=='}'&&!bracket.empty()&&bracket.top()=='{'){ bracket.pop(); }else{ return false; } } return bracket.empty(); } };
解题思路2:将可能出现的左右括号分别放进两个字符串里去,左右括号的位置相互对应,设立一个栈来存储左括号,然后使用find函数找到右括号在右括号字符串中的位置n后,取出左括号字符串位置n的字符,与栈顶元素比较。
代码2:
class Solution { public: bool isValid(string s) { stack<char> bracket; string left("([{"); string right(")]}"); for(int i=0;i<s.size();i++){ if(left.find(s[i])!=string::npos){ bracket.push(s[i]); }else if(!bracket.empty()&&bracket.top()==left[right.find(s[i])]){ bracket.pop(); }else{ return false; } } return bracket.empty(); } };