题目描述：
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
题源：here；完整代码：here
思路：
两个方案，第一个是消除法，即如果相邻的两个符号配对的话就删除它们，看最后是否有没有被删除的。第二个是堆栈法，我们用一个堆栈记录我们出现的符号，如果是左边的符号就压入堆栈，如果是右边的就看它和栈顶能不能配对：能配对就让栈顶元素出栈，不能就直接返回false。最后也需要查看堆栈是否为空，是的话返回true，否则返回false。
方案1
bool isValid_1(string s) {
if (s.size() % 2) return false;
map<char, char> pairs;
pairs.insert(pair<char, char>('{', '}'));
pairs.insert(pair<char, char>('[', ']'));
pairs.insert(pair<char, char>('(', ')'));
while (s.size()){
int idx = 0;
int s_len = s.size();
for (int i = 0; i < s_len-1; i++){
if (pairs[s[i]] == s[i + 1]){
s.erase(i, 2);
break;
}
}
if (s_len == s.size()) break;
}
if (s.size()) return false;
else return true;
}方案2
bool isValid(string s){
if (s.size() % 2) return false;
map<char, char> pairs;
pairs.insert(pair<char, char>('{', '}'));
pairs.insert(pair<char, char>('[', ']'));
pairs.insert(pair<char, char>('(', ')'));
stack<char> strs;
for (int i = 0; i < s.size(); i++){
if (s[i] == '{' || s[i] == '[' || s[i] == '('){
strs.push(s[i]);
}
else {
if (strs.empty()) return false;
if (pairs[strs.top()] == s[i] || pairs[strs.top()] == s[i] || pairs[strs.top()] == s[i]){
strs.pop();
}
else return false;
}
}
return strs.empty();
}