LEETCODE-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.
利用栈的push和pop过程，将”(“,”{“,”[“将所有括号左边部分push入栈，当出现”)”,”}”,”]”右半边的括号时与栈顶元素进行匹配（是否是一对）若是则pop栈顶元素，并进行下一位的匹配；若不同则直接返回false。

class Solution {
public:
    bool isValid(string s) {
        stack<int> No1stack;
    size_t chair = 0;
    int n = s.size();
    while (chair < n) {
        if ((s[chair] == '{') || (s[chair] == '[') || (s[chair] == '('))
            No1stack.push(s[chair]);
        else {
            if (No1stack.size() == 0)
                return false;
            char top = No1stack.top();
            switch (s[chair]) {
            case '}':
                if (top == '{')
                    No1stack.pop();
                else
                    return false;
                break;
            case ']':
                if (top == '[')
                    No1stack.pop();
                else
                    return false;
                break;
            case ')':
                if (top == '(')
                    No1stack.pop();
                else
                    return false;
                break;
            }
        }
        chair++;
    }
    if (No1stack.size() == 0)
        return true;
    else
            return false;
    }
};ring，它是原始string的一部分或全部的拷贝。

class Solution {
public:
bool isValid(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int len = s.length();
if(len==0) return true;
if(len%2) return false;
for(int i = 1; i < len; i++)
if(s[i-1] == ‘{’ && s[i]==’}’)
{
string ss = s.substr(0, i-1) + s.substr(i+1);
return isValid(ss);
}
else if(s[i-1] == ‘[’ && s[i]==’]’)
{
string ss = s.substr(0, i-1) + s.substr(i+1);
return isValid(ss);
}
else if(s[i-1] == ‘(’ && s[i]==’)’)
{
string ss = s.substr(0, i-1) + s.substr(i+1);
return isValid(ss);
}
return false;
}

};
“`