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:
- Any left parenthesis
'('must have a corresponding right parenthesis')'. - Any right parenthesis
')'must have a corresponding left parenthesis'('. - Left parenthesis
'('must go before the corresponding right parenthesis')'. '*'could be treated as a single right parenthesis')'or a single left parenthesis'('or an empty string.- An empty string is also valid.
Example 1:
Input: "()" Output: True
Example 2:
Input: "(*)" Output: True
Example 3:
Input: "(*))" Output: True
Note:
- 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:
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:
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:
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:
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
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++:
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++:
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++:
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