LeetCode Valid Parentheses

最新推荐文章于 2019-07-31 23:36:17 发布

转载最新推荐文章于 2019-07-31 23:36:17 发布 · 68 阅读

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/lailailai/p/3660043.html

本文提供了两种使用 C++ 实现的有效字符串括号匹配算法。一种是通过记录每种开括号的数量并检查闭括号是否与最近的开括号正确配对；另一种是利用哈希表来存储括号对，并使用栈来跟踪开括号。这两种方法都能有效地验证输入字符串中的括号是否按正确的顺序关闭。

 1 class Solution {
 2 public:
 3     bool isValid(string s) {
 4         char open[] = {'(', '[', '{'};
 5         char close[]= {')', ']', '}'};
 6         
 7         int count[3] = {0, 0, 0};
 8         
 9         vector<char> last_open;
10         int k;
11         for (int i=0; i<s.size(); i++) {
12             char cur = s[i];
13             for (k=0; k<3 && cur != open[k]; k++);
14             if (k < 3) {
15                 count[k]++;
16                 last_open.push_back(open[k]);
17                 continue;
18             }
19             
20             for (k=0; k<3 && cur != close[k]; k++);
21             if (k == 3) continue; // should not be happened
22             if (--count[k] < 0 || open[k] != last_open.back()) {
23                 return false;
24             } else {
25                 last_open.pop_back();
26             }
27         }
28         
29         return (last_open.size() == 0) && !(count[0] | count[1] | count[2]);
30     }
31 };

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();
    }
};

转载于:https://www.cnblogs.com/lailailai/p/3660043.html