［LeetCode］Valid Parentheses

解题思路：
边界条件：stack是空，取到rightSet里面的字符，false；或者遍历str结束后，stack不为空，也是false
前条件：leftSet, rightSet, matchPair
不变式：遍历s，找到没match的pair，false
结束条件：遍历s结束

class Solution {
public:
set<char> leftSet;
set<char> rightSet;
unordered_map<char, char> matchPair;

Solution(){
leftSet.insert('(');
leftSet.insert('[');
leftSet.insert('{');

rightSet.insert(')');
rightSet.insert(']');
rightSet.insert('}');

matchPair['('] = ')';
matchPair['['] = ']';
matchPair['{'] = '}';
}
bool isValid(string s) {
if (s.length() == 0)    return true;

stack<char> parentStack;
for (int i = 0; i < s.length(); ++i){
if (isInLeftSet(s[i])){
parentStack.push(s[i]);

}else if (isInRightSet(s[i])){
if(parentStack.empty()){
return false;
}

char temp = parentStack.top();
if (isMatch(temp, s[i])){
parentStack.pop();
}else{
return false;
}
}else{
printf("invalid input.");
}
}
if (!parentStack.empty()){
return false;
}

return true;
}

bool isMatch(char left, char right){
if (matchPair.find(left) != matchPair.end()){
return matchPair[left] == right;
}else{
printf("invalid input");
}
return false;
}

bool isInLeftSet(char c){

return (leftSet.find(c) != leftSet.end()) ?  true : false;
}

bool isInRightSet(char c){

return (rightSet.find(c) != rightSet.end()) ? true : false;
}
};