Description
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].
分析
题目的意思是:验证一个字符串是否是合法字符串。
- 这道题目我想到用了栈,但是没想到栈存放的是索引,在最后用来判断“*”能否消去s1里面的做括号,如果star的索引比s1的小,说明就不能当成右括号使用,直接返回false。
代码
class Solution {
public:
bool checkValidString(string s) {
stack<int> s1,star;
for(int i=0;i<s.length();i++){
if(s[i]=='('){
s1.push(i);
}else if(s[i]==')'){
if(s1.empty()&&star.empty()){
return false;
}
if(s1.empty()&&!star.empty()){
star.pop();
}else{
s1.pop();
}
}else{
star.push(i);
}
}
while(!s1.empty()&&!star.empty()){
if(s1.top()>star.top()){
return false;
}
star.pop();
s1.pop();
}
return s1.empty();
}
};
代码二(python)
- 设两个变量l和h,l最少左括号的情况,h表示最多左括号的情况,其它情况在[l, h]之间。
- 遍历字符串,遇到左括号时,l和h都自增1;当遇到右括号时,当l大于0时,l才自减1,否则保持为0(因为其它*情况可能会平衡掉),而h减1;
- 当遇到星号时,当l大于0时,l才自减1(当作右括号),而h自增1(当作左括号)。如果h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l是否为0。
class Solution:
def checkValidString(self, s: str) -> bool:
l=0
h=0
for ch in s:
if(ch=='('):
l+=1
h+=1
elif(ch==')'):
if(l>0):
l-=1
h-=1
else:
if(l>0):
l-=1
h+=1
if(h<0):
return False
return l==0
参考文献
[LeetCode] Valid Parenthesis String 验证括号字符串[LeetCode] 678. Valid Parenthesis String 验证括号字符串