Type：medium
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if:
Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example 1:
Input:"()"Output:true
Example 2:
Input:"()[]{}"Output:true
Example 3:
Input:"(]"Output:false
Example 4:
Input:"([)]"Output:false
Example 5:
Input:"{[]}"Output:true
此题题意为判断一个字符串是否为合法字符，与代码符号规则相同，左括号后面必须有配对的右括号，且同一种类型的左右括号之间必须没有单独的其他括号。
此题解法为开一个left数组，用来储存所有出现过的左括号，当出现右括号时，将它与左括号数组的最后一个左括号进行匹配，看是否为同一种括号，若是，则在数组中删去这个括号，继续遍历；否则返回 false。当遍历到最后数组为空，且没有多余的右括号，则返回 true 。
class Solution {
public:
bool isValid(string s) {
string left;
string right;
int L=s.length();
for(int i=0; i<L; i++){
if(s[i] == '(' || s[i] == '{' || s[i] == '[') left.push_back(s[i]);
else if(s[i] == ')' || s[i] == '}' || s[i] == ']') right.push_back(s[i]);
if(i>0 && right.back() == ')') {
if(left.back() == '('){
left.pop_back();
right.pop_back();
}
else return false;
}
else if(i>0 && right.back() == ']') {
if(left.back() == '['){
left.pop_back();
right.pop_back();
}
else return false;
}
else if(i>0 && right.back() == '}') {
if(left.back() == '{'){
left.pop_back();
right.pop_back();
}
else return false;
}
}
if(left.length() != 0) return false;
else if(right.length() != 0) return false;
else return true;
}
};
