LeetCode Valid Parentheses

class Solution {
public:
    bool isValid(string s) {
        char open[] = {'(', '[', '{'};
        char close[]= {')', ']', '}'};
        
        int count[3] = {0, 0, 0};
        
        vector<char> last_open;
        int k;
        for (int i=0; i<s.size(); i++) {
            char cur = s[i];
            for (k=0; k<3 && cur != open[k]; k++);
            if (k < 3) {
                count[k]++;
                last_open.push_back(open[k]);
                continue;
            }
            
            for (k=0; k<3 && cur != close[k]; k++);
            if (k == 3) continue; // should not be happened
            if (--count[k] < 0 || open[k] != last_open.back()) {
                return false;
            } else {
                last_open.pop_back();
            }
        }
        
        return (last_open.size() == 0) && !(count[0] | count[1] | count[2]);
    }
};

再水一发！

第二轮：

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.

class Solution {
public:
    bool isValid(string s) {
        char pair[256] = {0};
        pair['{'] = '}';
        pair['('] = ')';
        pair['['] = ']';
        stack<char> seq;
        int len = s.size();
        for (int i=0; i<len; i++) {
            char cur = s[i];
            if (cur == '(' || cur == '{' || cur == '[') {
                seq.push(cur);
                continue;
            }
            if (seq.empty()) {
                return false;
            } else {
                if (pair[seq.top()] != cur) {
                    return false;
                }
                seq.pop();
            }
        }
        return seq.empty()
    }
};

不够仔细啊，参考题解：

class Solution {
public:
    bool isValid(string s) {
        char pair[256] = {0};
        pair['{'] = '}';
        pair['('] = ')';
        pair['['] = ']';
        stack<char> seq;
        int len = s.size();
        for (int i=0; i<len; i++) {
            char cur = s[i];
            if (pair[cur] != 0) {
                seq.push(cur);
                continue;
            } else if (seq.empty() || pair[seq.top()] != cur) {
                return false;
            }
            seq.pop();
        }
        return seq.empty();
    }
};