[LeetCode] 678. Valid Parenthesis String 验证括号字符串

Given a string containing only three types of characters: '(', ')' and '*', write a function to check whether this string is valid. We define the validity of a string by these rules:

  1. Any left parenthesis '(' must have a corresponding right parenthesis ')'.
  2. Any right parenthesis ')' must have a corresponding left parenthesis '('.
  3. Left parenthesis '(' must go before the corresponding right parenthesis ')'.
  4. '*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string.
  5. An empty string is also valid.

Example 1:

Input: "()"
Output: True

Example 2:

Input: "(*)"
Output: True

Example 3:

Input: "(*))"
Output: True

Note:

  1. The string size will be in the range [1, 100].

判断给定的字符串的括号匹配是否合法， 20. Valid Parentheses 的变形，这题里只有小括号'()'和*，*号可以代表'(', ')'或者相当于没有。

解法1: 迭代

解法2: 递归，最容易想到的解法，用一个变量cnt记录左括号的数量，当遇到*号时分为三种情况递归下去。Python TLE,  C++: 1148 ms

解法2: 最优解法，设两个变量cmin和cmax，cmin最少左括号的情况，cmax表示最多左括号的情况，其它情况在[cmin, cmax]之间。遍历字符串，遇到左括号时，cmin和cmax都自增1；当遇到右括号时，当cmin大于0时，cmin才自减1，否则保持为0(因为其它*情况可能会平衡掉)，而cmax减1；当遇到星号时，当cmin大于0时，cmin才自减1(当作右括号)，而cmax自增1(当作左括号)。如果cmax小于0，说明到此字符时前面的右括号太多，有*也无法平衡掉，返回false。当循环结束后，返回cmin是否为0。

Java:

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public boolean checkValidString(String s) {
        int low = 0;
        int high = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') {
                low++;
                high++;
            } else if (s.charAt(i) == ')') {
                if (low > 0) {
                    low--;
                }
                high--;
            } else {
                if (low > 0) {
                    low--;
                }
                high++;
            }
            if (high < 0) {
                return false;
            }
        }
        return low == 0;
    }

Python:

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class Solution(object):
    def checkValidString(self, s):
        """
        :type s: str
        :rtype: bool
        """
        lower, upper = 0, 0  # keep lower bound and upper bound of '(' counts
        for c in s:
            lower += 1 if c == '(' else -1
            upper -= 1 if c == ')' else -1
            if upper < 0: break
            lower = max(lower, 0)
        return lower == 0  # range of '(' count is valid　　

Python:

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def checkValidString(self, s):
        cmin = cmax = 0
        for i in s:
            if i == '(':
                cmax += 1
                cmin += 1
            if i == ')':
                cmax -= 1
                cmin = max(cmin - 1, 0)
            if i == '*':
                cmax += 1
                cmin = max(cmin - 1, 0)
            if cmax < 0:
                return False
        return cmin == 0

Python:

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def checkValidString(self, s):
        cmin = cmax = 0
        for i in s:
            cmax = cmax-1 if i==')' else cmax+1
            cmin = cmin+1 if i=='(' else max(cmin - 1, 0)
            if cmax < 0: return False
        return cmin == 0　

Python: TLE

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class Solution(object):
    def checkValidString(self, s):
        """
        :type s: str
        :rtype: bool
        """
        return self.helper(s, 0)
         
    def helper(self, s, count):
        # if not s:
        #     return count == 0
        if count < 0:
            return False
 
        for i in xrange(len(s)):
            if s[i] == '(':
                count += 1
            elif s[i] == ')':
                if count <= 0:
                    return False
                else:
                    count -= 1
            else:
                return self.helper(s[i+1:], count+1)  or \
                       self.helper(s[i+1:], count)  or \
                       self.helper(s[i+1:], count-1)
             
        return count == 0   
 
     
if __name__ == '__main__':
    print Solution().checkValidString("(((((*(()((((*((**(((()()*)()()()*((((**)())*)*)))))))(())(()))())((*()()(((()((()*(())*(()**)()(())")　　

C++:

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class Solution {
public:
    bool checkValidString(string s) {
        stack<int> left, star;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == '*') star.push(i);
            else if (s[i] == '(') left.push(i);
            else {
                if (left.empty() && star.empty()) return false;
                if (!left.empty()) left.pop();
                else star.pop();
            }
        }
        while (!left.empty() && !star.empty()) {
            if (left.top() > star.top()) return false;
            left.pop(); star.pop();
        }
        return left.empty();
    }
};

C++:

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class Solution {
public:
    bool checkValidString(string s) {
        return helper(s, 0, 0);
    }
    bool helper(string s, int start, int cnt) {
        if (cnt < 0) return false;
        for (int i = start; i < s.size(); ++i) {
            if (s[i] == '(') {
                ++cnt;
            } else if (s[i] == ')') {
                if (cnt <= 0) return false;
                --cnt;
            } else {
                return helper(s, i + 1, cnt) || helper(s, i + 1, cnt + 1) || helper(s, i + 1, cnt - 1);
            }
        }
        return cnt == 0;
    }
};

C++:

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class Solution {
public:
    bool checkValidString(string s) {
        int low = 0, high = 0;
        for (char c : s) {
            if (c == '(') {
                ++low; ++high;
            } else if (c == ')') {
                if (low > 0) --low;
                --high;
            } else {
                if (low > 0) --low;
                ++high;
            }
            if (high < 0) return false;
        }
        return low == 0;
    }
};

　　

　　

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[LeetCode] 20. Valid Parentheses 合法括号

 

　　

 

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